assignment 6.f.o.i.l. and special cases

6.3 Multiply Polynomials

Learning objectives.

By the end of this section, you will be able to:

• Multiply a polynomial by a monomial
• Multiply a binomial by a binomial
• Multiply a trinomial by a binomial

Be Prepared 6.3

Before you get started, take this readiness quiz.

• Distribute: 2 ( x + 3 ) . 2 ( x + 3 ) . If you missed this problem, review Example 1.132 .
• Combine like terms: x 2 + 9 x + 7 x + 63 . x 2 + 9 x + 7 x + 63 . If you missed this problem, review Example 1.24 .

Multiply a Polynomial by a Monomial

We have used the Distributive Property to simplify expressions like 2 ( x − 3 ) 2 ( x − 3 ) . You multiplied both terms in the parentheses, x and 3 x and 3 , by 2, to get 2 x − 6 2 x − 6 . With this chapter’s new vocabulary, you can say you were multiplying a binomial, x − 3 x − 3 , by a monomial, 2.

Multiplying a binomial by a monomial is nothing new for you! Here’s an example:

Example 6.28

Multiply: 4 ( x + 3 ) . 4 ( x + 3 ) .

Try It 6.55

Multiply: 5 ( x + 7 ) . 5 ( x + 7 ) .

Try It 6.56

Multiply: 3 ( y + 13 ) . 3 ( y + 13 ) .

Example 6.29

Multiply: y ( y − 2 ) . y ( y − 2 ) .

Try It 6.57

Multiply: x ( x − 7 ) . x ( x − 7 ) .

Try It 6.58

Multiply: d ( d − 11 ) . d ( d − 11 ) .

Example 6.30

Multiply: 7 x ( 2 x + y ) . 7 x ( 2 x + y ) .

Try It 6.59

Multiply: 5 x ( x + 4 y ) . 5 x ( x + 4 y ) .

Try It 6.60

Multiply: 2 p ( 6 p + r ) . 2 p ( 6 p + r ) .

Example 6.31

Multiply: −2 y ( 4 y 2 + 3 y − 5 ) . −2 y ( 4 y 2 + 3 y − 5 ) .

Try It 6.61

Multiply: −3 y ( 5 y 2 + 8 y − 7 ) . −3 y ( 5 y 2 + 8 y − 7 ) .

Try It 6.62

Multiply: 4 x 2 ( 2 x 2 − 3 x + 5 ) . 4 x 2 ( 2 x 2 − 3 x + 5 ) .

Example 6.32

Multiply: 2 x 3 ( x 2 − 8 x + 1 ) . 2 x 3 ( x 2 − 8 x + 1 ) .

Try It 6.63

Multiply: 4 x ( 3 x 2 − 5 x + 3 ) . 4 x ( 3 x 2 − 5 x + 3 ) .

Try It 6.64

Multiply: −6 a 3 ( 3 a 2 − 2 a + 6 ) . −6 a 3 ( 3 a 2 − 2 a + 6 ) .

Example 6.33

Multiply: ( x + 3 ) p . ( x + 3 ) p .

Try It 6.65

Multiply: ( x + 8 ) p . ( x + 8 ) p .

Try It 6.66

Multiply: ( a + 4 ) p . ( a + 4 ) p .

Multiply a Binomial by a Binomial

Just like there are different ways to represent multiplication of numbers, there are several methods that can be used to multiply a binomial times a binomial. We will start by using the Distributive Property.

Multiply a Binomial by a Binomial Using the Distributive Property

Look at Example 6.33 , where we multiplied a binomial by a monomial .

Notice that before combining like terms, you had four terms. You multiplied the two terms of the first binomial by the two terms of the second binomial—four multiplications.

Example 6.34

Multiply: ( y + 5 ) ( y + 8 ) . ( y + 5 ) ( y + 8 ) .

Try It 6.67

Multiply: ( x + 8 ) ( x + 9 ) . ( x + 8 ) ( x + 9 ) .

Try It 6.68

Multiply: ( 5 x + 9 ) ( 4 x + 3 ) . ( 5 x + 9 ) ( 4 x + 3 ) .

Example 6.35

Multiply: ( 2 y + 5 ) ( 3 y + 4 ) . ( 2 y + 5 ) ( 3 y + 4 ) .

Try It 6.69

Multiply: ( 3 b + 5 ) ( 4 b + 6 ) . ( 3 b + 5 ) ( 4 b + 6 ) .

Try It 6.70

Multiply: ( a + 10 ) ( a + 7 ) . ( a + 10 ) ( a + 7 ) .

Example 6.36

Multiply: ( 4 y + 3 ) ( 2 y − 5 ) . ( 4 y + 3 ) ( 2 y − 5 ) .

Try It 6.71

Multiply: ( 5 y + 2 ) ( 6 y − 3 ) . ( 5 y + 2 ) ( 6 y − 3 ) .

Try It 6.72

Multiply: ( 3 c + 4 ) ( 5 c − 2 ) . ( 3 c + 4 ) ( 5 c − 2 ) .

Example 6.37

Multiply: ( x - 2 ) ( x − y ) . ( x - 2 ) ( x − y ) .

Try It 6.73

Multiply: ( a + 7 ) ( a − b ) . ( a + 7 ) ( a − b ) .

Try It 6.74

Multiply: ( x + 5 ) ( x − y ) . ( x + 5 ) ( x − y ) .

Multiply a Binomial by a Binomial Using the FOIL Method

Remember that when you multiply a binomial by a binomial you get four terms. Sometimes you can combine like terms to get a trinomial , but sometimes, like in Example 6.37 , there are no like terms to combine.

Let’s look at the last example again and pay particular attention to how we got the four terms.

Where did the first term, x 2 x 2 , come from?

We abbreviate “First, Outer, Inner, Last” as FOIL. The letters stand for ‘ F irst, O uter, I nner, L ast’. The word FOIL is easy to remember and ensures we find all four products.

Let’s look at ( x + 3 ) ( x + 7 ) ( x + 3 ) ( x + 7 ) .

Notice how the terms in third line fit the FOIL pattern.

Now we will do an example where we use the FOIL pattern to multiply two binomials.

Example 6.38

How to multiply a binomial by a binomial using the foil method.

Multiply using the FOIL method: ( x + 5 ) ( x + 9 ) . ( x + 5 ) ( x + 9 ) .

Try It 6.75

Multiply using the FOIL method: ( x + 6 ) ( x + 8 ) . ( x + 6 ) ( x + 8 ) .

Try It 6.76

Multiply using the FOIL method: ( y + 17 ) ( y + 3 ) . ( y + 17 ) ( y + 3 ) .

We summarize the steps of the FOIL method below. The FOIL method only applies to multiplying binomials, not other polynomials!

Multiply two binomials using the FOIL method

When you multiply by the FOIL method, drawing the lines will help your brain focus on the pattern and make it easier to apply.

Example 6.39

Multiply: ( y − 7 ) ( y + 4 ) . ( y − 7 ) ( y + 4 ) .

Try It 6.77

Multiply: ( x − 7 ) ( x + 5 ) . ( x − 7 ) ( x + 5 ) .

Try It 6.78

Multiply: ( b − 3 ) ( b + 6 ) . ( b − 3 ) ( b + 6 ) .

Example 6.40

Multiply: ( 4 x + 3 ) ( 2 x − 5 ) . ( 4 x + 3 ) ( 2 x − 5 ) .

Try It 6.79

Multiply: ( 3 x + 7 ) ( 5 x − 2 ) . ( 3 x + 7 ) ( 5 x − 2 ) .

Try It 6.80

Multiply: ( 4 y + 5 ) ( 4 y − 10 ) . ( 4 y + 5 ) ( 4 y − 10 ) .

The final products in the last four examples were trinomials because we could combine the two middle terms. This is not always the case.

Example 6.41

Multiply: ( 3 x − y ) ( 2 x − 5 ) . ( 3 x − y ) ( 2 x − 5 ) .

Try It 6.81

Multiply: ( 10 c − d ) ( c − 6 ) . ( 10 c − d ) ( c − 6 ) .

Try It 6.82

Multiply: ( 7 x − y ) ( 2 x − 5 ) . ( 7 x − y ) ( 2 x − 5 ) .

Be careful of the exponents in the next example.

Example 6.42

Multiply: ( n 2 + 4 ) ( n − 1 ) . ( n 2 + 4 ) ( n − 1 ) .

Try It 6.83

Multiply: ( x 2 + 6 ) ( x − 8 ) . ( x 2 + 6 ) ( x − 8 ) .

Try It 6.84

Multiply: ( y 2 + 7 ) ( y − 9 ) . ( y 2 + 7 ) ( y − 9 ) .

Example 6.43

Multiply: ( 3 p q + 5 ) ( 6 p q − 11 ) . ( 3 p q + 5 ) ( 6 p q − 11 ) .

Try It 6.85

Multiply: ( 2 a b + 5 ) ( 4 a b − 4 ) . ( 2 a b + 5 ) ( 4 a b − 4 ) .

Try It 6.86

Multiply: ( 2 x y + 3 ) ( 4 x y − 5 ) . ( 2 x y + 3 ) ( 4 x y − 5 ) .

Multiply a Binomial by a Binomial Using the Vertical Method

The FOIL method is usually the quickest method for multiplying two binomials, but it only works for binomials. You can use the Distributive Property to find the product of any two polynomials. Another method that works for all polynomials is the Vertical Method. It is very much like the method you use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.

Now we’ll apply this same method to multiply two binomials.

Example 6.44

Multiply using the Vertical Method: ( 3 y − 1 ) ( 2 y − 6 ) . ( 3 y − 1 ) ( 2 y − 6 ) .

It does not matter which binomial goes on the top.

Multiply 3 y − 1 by −6 . Multiply 3 y − 1 by 2y. Add like terms. 3 y − 1 × 2 y − 6 ________ −18 y + 6 6 y 2 − 2 y  _____________ 6 y 2 − 20 y + 6 partial product partial product product Multiply 3 y − 1 by −6 . Multiply 3 y − 1 by 2y. Add like terms. 3 y − 1 × 2 y − 6 ________ −18 y + 6 6 y 2 − 2 y  _____________ 6 y 2 − 20 y + 6 partial product partial product product

Notice the partial products are the same as the terms in the FOIL method.

Try It 6.87

Multiply using the Vertical Method: ( 5 m − 7 ) ( 3 m − 6 ) . ( 5 m − 7 ) ( 3 m − 6 ) .

Try It 6.88

Multiply using the Vertical Method: ( 6 b − 5 ) ( 7 b − 3 ) . ( 6 b − 5 ) ( 7 b − 3 ) .

We have now used three methods for multiplying binomials. Be sure to practice each method, and try to decide which one you prefer. The methods are listed here all together, to help you remember them.

Multiplying Two Binomials

To multiply binomials, use the:

• Distributive Property
• FOIL Method
• Vertical Method

Remember, FOIL only works when multiplying two binomials.

Multiply a Trinomial by a Binomial

We have multiplied monomials by monomials, monomials by polynomials, and binomials by binomials. Now we’re ready to multiply a trinomial by a binomial . Remember, FOIL will not work in this case, but we can use either the Distributive Property or the Vertical Method. We first look at an example using the Distributive Property.

Example 6.45

Multiply using the Distributive Property: ( b + 3 ) ( 2 b 2 − 5 b + 8 ) . ( b + 3 ) ( 2 b 2 − 5 b + 8 ) .

Try It 6.89

Multiply using the Distributive Property: ( y − 3 ) ( y 2 − 5 y + 2 ) . ( y − 3 ) ( y 2 − 5 y + 2 ) .

Try It 6.90

Multiply using the Distributive Property: ( x + 4 ) ( 2 x 2 − 3 x + 5 ) . ( x + 4 ) ( 2 x 2 − 3 x + 5 ) .

Now let’s do this same multiplication using the Vertical Method.

Example 6.46

Multiply using the Vertical Method: ( b + 3 ) ( 2 b 2 − 5 b + 8 ) . ( b + 3 ) ( 2 b 2 − 5 b + 8 ) .

It is easier to put the polynomial with fewer terms on the bottom because we get fewer partial products this way.

Try It 6.91

Multiply using the Vertical Method: ( y − 3 ) ( y 2 − 5 y + 2 ) . ( y − 3 ) ( y 2 − 5 y + 2 ) .

Try It 6.92

Multiply using the Vertical Method: ( x + 4 ) ( 2 x 2 − 3 x + 5 ) . ( x + 4 ) ( 2 x 2 − 3 x + 5 ) .

We have now seen two methods you can use to multiply a trinomial by a binomial. After you practice each method, you’ll probably find you prefer one way over the other. We list both methods are listed here, for easy reference.

Multiplying a Trinomial by a Binomial

To multiply a trinomial by a binomial, use the:

Access these online resources for additional instruction and practice with multiplying polynomials:

• Multiplying Exponents 1
• Multiplying Exponents 2
• Multiplying Exponents 3

Section 6.3 Exercises

Practice makes perfect.

In the following exercises, multiply.

4 ( w + 10 ) 4 ( w + 10 )

6 ( b + 8 ) 6 ( b + 8 )

−3 ( a + 7 ) −3 ( a + 7 )

−5 ( p + 9 ) −5 ( p + 9 )

2 ( x − 7 ) 2 ( x − 7 )

7 ( y − 4 ) 7 ( y − 4 )

−3 ( k − 4 ) −3 ( k − 4 )

−8 ( j − 5 ) −8 ( j − 5 )

q ( q + 5 ) q ( q + 5 )

k ( k + 7 ) k ( k + 7 )

− b ( b + 9 ) − b ( b + 9 )

− y ( y + 3 ) − y ( y + 3 )

− x ( x − 10 ) − x ( x − 10 )

− p ( p − 15 ) − p ( p − 15 )

6 r ( 4 r + s ) 6 r ( 4 r + s )

5 c ( 9 c + d ) 5 c ( 9 c + d )

12 x ( x − 10 ) 12 x ( x − 10 )

9 m ( m − 11 ) 9 m ( m − 11 )

−9 a ( 3 a + 5 ) −9 a ( 3 a + 5 )

−4 p ( 2 p + 7 ) −4 p ( 2 p + 7 )

3 ( p 2 + 10 p + 25 ) 3 ( p 2 + 10 p + 25 )

6 ( y 2 + 8 y + 16 ) 6 ( y 2 + 8 y + 16 )

−8 x ( x 2 + 2 x − 15 ) −8 x ( x 2 + 2 x − 15 )

−5 t ( t 2 + 3 t − 18 ) −5 t ( t 2 + 3 t − 18 )

5 q 3 ( q 3 − 2 q + 6 ) 5 q 3 ( q 3 − 2 q + 6 )

4 x 3 ( x 4 − 3 x + 7 ) 4 x 3 ( x 4 − 3 x + 7 )

−8 y ( y 2 + 2 y − 15 ) −8 y ( y 2 + 2 y − 15 )

−5 m ( m 2 + 3 m − 18 ) −5 m ( m 2 + 3 m − 18 )

5 q 3 ( q 2 − 2 q + 6 ) 5 q 3 ( q 2 − 2 q + 6 )

9 r 3 ( r 2 − 3 r + 5 ) 9 r 3 ( r 2 − 3 r + 5 )

−4 z 2 ( 3 z 2 + 12 z − 1 ) −4 z 2 ( 3 z 2 + 12 z − 1 )

−3 x 2 ( 7 x 2 + 10 x − 1 ) −3 x 2 ( 7 x 2 + 10 x − 1 )

( 2 m − 9 ) m ( 2 m − 9 ) m

( 8 j − 1 ) j ( 8 j − 1 ) j

( w − 6 ) · 8 ( w − 6 ) · 8

( k − 4 ) · 5 ( k − 4 ) · 5

4 ( x + 10 ) 4 ( x + 10 )

6 ( y + 8 ) 6 ( y + 8 )

15 ( r − 24 ) 15 ( r − 24 )

12 ( v − 30 ) 12 ( v − 30 )

−3 ( m + 11 ) −3 ( m + 11 )

−4 ( p + 15 ) −4 ( p + 15 )

−8 ( z − 5 ) −8 ( z − 5 )

−3 ( x − 9 ) −3 ( x − 9 )

u ( u + 5 ) u ( u + 5 )

q ( q + 7 ) q ( q + 7 )

n ( n 2 − 3 n ) n ( n 2 − 3 n )

s ( s 2 − 6 s ) s ( s 2 − 6 s )

6 x ( 4 x + y ) 6 x ( 4 x + y )

5 a ( 9 a + b ) 5 a ( 9 a + b )

5 p ( 11 p − 5 q ) 5 p ( 11 p − 5 q )

12 u ( 3 u − 4 v ) 12 u ( 3 u − 4 v )

3 ( v 2 + 10 v + 25 ) 3 ( v 2 + 10 v + 25 )

6 ( x 2 + 8 x + 16 ) 6 ( x 2 + 8 x + 16 )

2 n ( 4 n 2 − 4 n + 1 ) 2 n ( 4 n 2 − 4 n + 1 )

3 r ( 2 r 2 − 6 r + 2 ) 3 r ( 2 r 2 − 6 r + 2 )

( 2 y − 9 ) y ( 2 y − 9 ) y

( 8 b − 1 ) b ( 8 b − 1 ) b

In the following exercises, multiply the following binomials using: ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.

( w + 5 ) ( w + 7 ) ( w + 5 ) ( w + 7 )

( y + 9 ) ( y + 3 ) ( y + 9 ) ( y + 3 )

( p + 11 ) ( p − 4 ) ( p + 11 ) ( p − 4 )

( q + 4 ) ( q − 8 ) ( q + 4 ) ( q − 8 )

In the following exercises, multiply the binomials. Use any method.

( x + 8 ) ( x + 3 ) ( x + 8 ) ( x + 3 )

( y + 7 ) ( y + 4 ) ( y + 7 ) ( y + 4 )

( y − 6 ) ( y − 2 ) ( y − 6 ) ( y − 2 )

( x − 7 ) ( x − 2 ) ( x − 7 ) ( x − 2 )

( w − 4 ) ( w + 7 ) ( w − 4 ) ( w + 7 )

( q − 5 ) ( q + 8 ) ( q − 5 ) ( q + 8 )

( p + 12 ) ( p − 5 ) ( p + 12 ) ( p − 5 )

( m + 11 ) ( m − 4 ) ( m + 11 ) ( m − 4 )

( 6 p + 5 ) ( p + 1 ) ( 6 p + 5 ) ( p + 1 )

( 7 m + 1 ) ( m + 3 ) ( 7 m + 1 ) ( m + 3 )

( 2 t − 9 ) ( 10 t + 1 ) ( 2 t − 9 ) ( 10 t + 1 )

( 3 r − 8 ) ( 11 r + 1 ) ( 3 r − 8 ) ( 11 r + 1 )

( 5 x − y ) ( 3 x − 6 ) ( 5 x − y ) ( 3 x − 6 )

( 10 a − b ) ( 3 a − 4 ) ( 10 a − b ) ( 3 a − 4 )

( a + b ) ( 2 a + 3 b ) ( a + b ) ( 2 a + 3 b )

( r + s ) ( 3 r + 2 s ) ( r + s ) ( 3 r + 2 s )

( 4 z − y ) ( z − 6 ) ( 4 z − y ) ( z − 6 )

( 5 x − y ) ( x − 4 ) ( 5 x − y ) ( x − 4 )

( x 2 + 3 ) ( x + 2 ) ( x 2 + 3 ) ( x + 2 )

( y 2 − 4 ) ( y + 3 ) ( y 2 − 4 ) ( y + 3 )

( x 2 + 8 ) ( x 2 − 5 ) ( x 2 + 8 ) ( x 2 − 5 )

( y 2 − 7 ) ( y 2 − 4 ) ( y 2 − 7 ) ( y 2 − 4 )

( 5 a b − 1 ) ( 2 a b + 3 ) ( 5 a b − 1 ) ( 2 a b + 3 )

( 2 x y + 3 ) ( 3 x y + 2 ) ( 2 x y + 3 ) ( 3 x y + 2 )

( 6 p q − 3 ) ( 4 p q − 5 ) ( 6 p q − 3 ) ( 4 p q − 5 )

( 3 r s − 7 ) ( 3 r s − 4 ) ( 3 r s − 7 ) ( 3 r s − 4 )

In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.

( x + 5 ) ( x 2 + 4 x + 3 ) ( x + 5 ) ( x 2 + 4 x + 3 )

( u + 4 ) ( u 2 + 3 u + 2 ) ( u + 4 ) ( u 2 + 3 u + 2 )

( y + 8 ) ( 4 y 2 + y − 7 ) ( y + 8 ) ( 4 y 2 + y − 7 )

( a + 10 ) ( 3 a 2 + a − 5 ) ( a + 10 ) ( 3 a 2 + a − 5 )

In the following exercises, multiply. Use either method.

( w − 7 ) ( w 2 − 9 w + 10 ) ( w − 7 ) ( w 2 − 9 w + 10 )

( p − 4 ) ( p 2 − 6 p + 9 ) ( p − 4 ) ( p 2 − 6 p + 9 )

( 3 q + 1 ) ( q 2 − 4 q − 5 ) ( 3 q + 1 ) ( q 2 − 4 q − 5 )

( 6 r + 1 ) ( r 2 − 7 r − 9 ) ( 6 r + 1 ) ( r 2 − 7 r − 9 )

Mixed Practice

( 10 y − 6 ) + ( 4 y − 7 ) ( 10 y − 6 ) + ( 4 y − 7 )

( 15 p − 4 ) + ( 3 p − 5 ) ( 15 p − 4 ) + ( 3 p − 5 )

( x 2 − 4 x − 34 ) − ( x 2 + 7 x − 6 ) ( x 2 − 4 x − 34 ) − ( x 2 + 7 x − 6 )

( j 2 − 8 j − 27 ) − ( j 2 + 2 j − 12 ) ( j 2 − 8 j − 27 ) − ( j 2 + 2 j − 12 )

5 q ( 3 q 2 − 6 q + 11 ) 5 q ( 3 q 2 − 6 q + 11 )

8 t ( 2 t 2 − 5 t + 6 ) 8 t ( 2 t 2 − 5 t + 6 )

( s − 7 ) ( s + 9 ) ( s − 7 ) ( s + 9 )

( x − 5 ) ( x + 13 ) ( x − 5 ) ( x + 13 )

( y 2 − 2 y ) ( y + 1 ) ( y 2 − 2 y ) ( y + 1 )

( a 2 − 3 a ) ( 4 a + 5 ) ( a 2 − 3 a ) ( 4 a + 5 )

( 3 n − 4 ) ( n 2 + n − 7 ) ( 3 n − 4 ) ( n 2 + n − 7 )

( 6 k − 1 ) ( k 2 + 2 k − 4 ) ( 6 k − 1 ) ( k 2 + 2 k − 4 )

( 7 p + 10 ) ( 7 p − 10 ) ( 7 p + 10 ) ( 7 p − 10 )

( 3 y + 8 ) ( 3 y − 8 ) ( 3 y + 8 ) ( 3 y − 8 )

( 4 m 2 − 3 m − 7 ) m 2 ( 4 m 2 − 3 m − 7 ) m 2

( 15 c 2 − 4 c + 5 ) c 4 ( 15 c 2 − 4 c + 5 ) c 4

( 5 a + 7 b ) ( 5 a + 7 b ) ( 5 a + 7 b ) ( 5 a + 7 b )

( 3 x − 11 y ) ( 3 x − 11 y ) ( 3 x − 11 y ) ( 3 x − 11 y )

( 4 y + 12 z ) ( 4 y − 12 z ) ( 4 y + 12 z ) ( 4 y − 12 z )

Everyday Math

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 13 times 15. Think of 13 as 10 + 3 10 + 3 and 15 as 10 + 5 10 + 5 .

• ⓐ Multiply ( 10 + 3 ) ( 10 + 5 ) ( 10 + 3 ) ( 10 + 5 ) by the FOIL method.
• ⓑ Multiply 13 · 15 13 · 15 without using a calculator.
• ⓒ Which way is easier for you? Why?

Mental math You can use binomial multiplication to multiply numbers without a calculator. Say you need to multiply 18 times 17. Think of 18 as 20 − 2 20 − 2 and 17 as 20 − 3 20 − 3 .

• ⓐ Multiply ( 20 − 2 ) ( 20 − 3 ) ( 20 − 2 ) ( 20 − 3 ) by the FOIL method.
• ⓑ Multiply 18 · 17 18 · 17 without using a calculator.

Writing Exercises

Which method do you prefer to use when multiplying two binomials: the Distributive Property, the FOIL method, or the Vertical Method? Why?

Which method do you prefer to use when multiplying a trinomial by a binomial: the Distributive Property or the Vertical Method? Why?

Multiply the following:

( x + 2 ) ( x − 2 ) ( y + 7 ) ( y − 7 ) ( w + 5 ) ( w − 5 ) ( x + 2 ) ( x − 2 ) ( y + 7 ) ( y − 7 ) ( w + 5 ) ( w − 5 )

Explain the pattern that you see in your answers.

( m − 3 ) ( m + 3 ) ( n − 10 ) ( n + 10 ) ( p − 8 ) ( p + 8 ) ( m − 3 ) ( m + 3 ) ( n − 10 ) ( n + 10 ) ( p − 8 ) ( p + 8 )

( p + 3 ) ( p + 3 ) ( q + 6 ) ( q + 6 ) ( r + 1 ) ( r + 1 ) ( p + 3 ) ( p + 3 ) ( q + 6 ) ( q + 6 ) ( r + 1 ) ( r + 1 )

( x − 4 ) ( x − 4 ) ( y − 1 ) ( y − 1 ) ( z − 7 ) ( z − 7 ) ( x − 4 ) ( x − 4 ) ( y − 1 ) ( y − 1 ) ( z − 7 ) ( z − 7 )

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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• Book title: Elementary Algebra
• Publication date: Feb 22, 2017
• Location: Houston, Texas
• Book URL: https://openstax.org/books/elementary-algebra/pages/1-introduction
• Section URL: https://openstax.org/books/elementary-algebra/pages/6-3-multiply-polynomials

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Multiplying Binomials: " foil " (and a warning)

Simple FOIL Longer Polys

What is " foil "?

foil is a special method of polynomial multiplication, useful ONLY for a two-term polynomial times another two-term polynomial. The letters f-o-i-l come from the words "first", "outer", "inner", "last", and are a memory device for helping you remember how to multiply horizontally, without having to write out the distribution like I did on the previous page , and without dropping any terms or signs.

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MathHelp.com

The FOIL Method

Below is a graphic displaying what foil stands for:

In other words, foil tells you to multiply the first terms in each of the parentheses, then multiply the two terms that are on the "outside" (that is, the two terms which are furthest from each other), then the two terms that are on the "inside" (that is, the two terms which are closest to each other), and then the last terms in each of the parentheses.

Using the example from the previous page, foil works like this:

• Use foil to simplify ( x  + 3)( x  + 2)

Okay; the instructions specify the method I have to use, so here goes:

"first": ( x )( x ) = x 2

"outer": ( x )(+2) = +2 x

"inner": (+3)( x ) = 3 x

"last": (+3)(+2) = +6

Adding the results of the four multiplications together, and combining the two "like" terms in the middle, I get:

x 2 + 2 x + 3 x + 6

x 2 + 5 x + 6

This is the same answer as we got when we multiplied horizontally and vertically.

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Many instructors in later math classes come to hate " foil " because it often seems to serve mostly to confuse students when they reach more advanced material. Unfortunately, foil tends to be taught in earlier algebra courses as "the" way to multiply all polynomials, which is clearly not true. (As soon as either one of the polynomials has more than a "first" and "last" term in its parentheses, you're hosed if you try to use F foil , because those terms won't "fit".)

And foil is, essentially, just a means of keeping track of what you're doing when you're multiplying horizontally. But, for multiplications of larger numbers, you already know that vertical is the way to go. It's the same in algebra. When multiplying larger polynomials, just about everybody switches to vertical multiplication; it's just so much easier to use.

If you want to use foil , that's fine, but (warning!) keep its restriction in mind: you can ONLY use it for the special case of multiplying two binomials. You can NOT use it at ANY other time!

• Simplify ( x  − 4)( x  − 3)

The instructions don't tell me what method I have to use for this multiplication, so I'll go vertical:

So my answer is:

x 2 − 7 x + 12

Using foil would give the same answer:

"outer": ( x )(−3) = −3 x

"inner": (−4)( x ) = −4 x

"last": (−4)(−3) = +12

result: ( x 2 ) + (−3 x ) + (−4 x ) + (+12)

But multiplying vertically takes less time, space, and care than does multiplying with foil . When I first encountered vertical multiplication of polynomials, I immediately switched to this method.

• Multiply and simplify: ( x  − 3 y )( x  +  y )

This multiplication looks like it might be more complicated, because of the second variable, but the process doesn't care how many terms have variables. So I can proceed as usual.

Vertical multiplication gives me this:

So my hand-in answer is:

x 2 − 2 xy − 3 y 2

Using foil would give me:

"outer": ( x )( y ) = xy

"inner": (−3 y )( x ) = −3 xy

"last": (−3 y )( y ) = −3 y 2

result: ( x 2 ) + ( xy ) + (−3 xy ) + (−3 y 2 )

Multiply and simplify: (2 x  − 5)(3 x  + 4)

This multiplication has some slightly larger coefficients, but the process is the same. I'll work vertically:

answer: 6 x 2 − 7 x − 20

I won't bother doing the foil process for this exercise.

Multiply: (−4 x  − 1)(5 x  − 7)

The factors in this product have a lot of "minus" signs, but the process remains the same:

answer: −20 x 2 + 23 x + 7

Expand: ( x  − 3) 2

Some students will try to take a shortcut with this and, thinking that the power somehow "distributes" over the terms, they'll square each of the terms (getting an x 2 and a 9 , but no middle terms with x 's in them) and... then they'll be confused about what signs should go where.

Instead, I'm going to write out the square explicitly; the expression they gave me means the following:

( x − 3)( x − 3)

This multiplies as follows:

answer: x 2 − 6 + 9

Expand and simplify: katex.render("\\small{ \\bm{\\color{green}{ (x + \\frac{1}{3})(x + \\frac{1}{2}) }}}", typed01); ( x  + 1/3)( x  + 1/2)

Fractions? Really?!? Well, the process will work just the same with fractions as it did with whole numbers:

Off to the side on my scratch-paper, I add the fractions for the middle term:

1/2 + 1/3 = 3/6 + 2/6 = 5/6

So my simplified answer is:

x 2 + (5/6) x + 1/6

Before we move on, please allow me reiterate what I said at the beginning of this page: " foil " works ONLY for the specific and special case of a two-term expression times another two-term expression. It does NOT apply in ANY other case. You should not rely on foil for general multiplication, and should not expect it to "work" for every multiplication, or even for most multiplications. If you only learn foil , you will not have learned all you need to know, and this will cause you problems later on down the road.

Yes, I'm ranting. But I have seen too many students be greatly hampered in their studies by an unthinking over-reliance on foil . Their instructors often never even taught them any method for multiplying other sorts of polynomials.

For your own sake, take the time to read the next page and learn how to multiply general polynomials properly.

You can use the Mathway widget below to practice multiplying binomials. Try the entered exercise, or type in your own exercise. Then click the button and select "Multiply" or "Simplify" to compare your answer to Mathway's. (Or skip the widget, and continue with the lesson.)

Please accept "preferences" cookies in order to enable this widget.

(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)

URL: https://www.purplemath.com/modules/polymult2.htm

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Module 7: Polynomials and Polynomial Functions

7.4 – factoring special cases, learning objectives.

• Factoring perfect square trinomials
• Factoring s difference of squares
• Factoring the sum of cubes.

Factoring the difference of cubes

• Factoring by substitution

Factoring completely

• (7.4.4) – General factoring strategy

Why learn how to factor special cases?

Some people like to find patterns in the world around them, like a game.  There are some polynomials that, when factored, follow a specific pattern.

These include:

Perfect square trinomials of the form: [latex]{a}^{2}+2ab+{b}^{2}[/latex] or [latex]{a}^{2}-2ab+{b}^{2}[/latex]

A difference of squares: [latex]{a}^{2}-{b}^{2}[/latex]

A sum of cubes: [latex]{a}^{3}+{b}^{3}[/latex]

A difference of cubes: [latex]{a}^{3}-{b}^{3}[/latex]

In this lesson you will see you can factor each of these types of polynomials following a specific pattern.  You will also learn how to factor polynomials that have negative or fractional exponents.

Some people find it helpful to know when they can take a shortcut to avoid doing extra work.  There are some polynomials that will always factor a certain way, and for those we offer a shortcut.  Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares.  The most important skill you will use in this section will be recognizing when you can use the shortcuts.

(7.4.1) – Factoring special cases – squares

Factoring perfect square trinomials.

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:

In the following example we will show you how to define a, and b so you can use the shortcut.

Factor [latex]25{x}^{2}+20x+4[/latex].

First, notice that [latex]25{x}^{2}[/latex] and [latex]4[/latex] are perfect squares because [latex]25{x}^{2}={\left(5x\right)}^{2}[/latex] and [latex]4={2}^{2}[/latex].

This means that [latex]a=5x\text{ and }b=2[/latex]

Next, check to see if the middle term is equal to [latex]2ab[/latex], which it is:

[latex]2ab = 2\left(5x\right)\left(2\right)=20x[/latex].

Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a+b\right)}^{2}={\left(5x+2\right)}^{2}[/latex].

[latex]25{x}^{2}+20x+4={\left(5x+2\right)}^{2}[/latex]

In the next example, we will show that we can use [latex]1 = 1^2[/latex] to factor a polynomial with a term equal to 1.

Factor [latex]49{x}^{2}-14x+1[/latex].

First, notice that [latex]49{x}^{2}[/latex] and [latex]1[/latex] are perfect squares because [latex]49{x}^{2}={\left(7x\right)}^{2}[/latex] and [latex]1={1}^{2}[/latex].

This means that [latex]a=7x[/latex], we could say that [latex]b=1[/latex], but would that give a middle term of [latex]-14x[/latex]? We will need to choose [latex]b = -1[/latex] to get the results we want:

[latex]2ab = 2\left(7x\right)\left(-1\right)=-14x[/latex].

Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(a-b\right)}^{2}={\left(7x-1\right)}^{2}[/latex].

[latex]49{x}^{2}-14x+1={\left(7x-1\right)}^{2}[/latex]

In the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula.

We can summarize our process in the following way:

Given a perfect square trinomial, factor it into the square of a binomial.

• Confirm that the first and last term are perfect squares.
• Confirm that the middle term is twice the product of [latex]ab[/latex].
• Write the factored form as [latex]{\left(a+b\right)}^{2}[/latex], or [latex]{\left(a-b\right)}^{2}[/latex].

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

Factor [latex]9{x}^{2}-25[/latex].

Notice that [latex]9{x}^{2}[/latex] and [latex]25[/latex] are perfect squares because [latex]9{x}^{2}={\left(3x\right)}^{2}[/latex] and [latex]25={5}^{2}[/latex].

This means that [latex]a=3x,\text{ and }b=5[/latex]

The polynomial represents a difference of squares and can be rewritten as [latex]\left(3x+5\right)\left(3x - 5\right)[/latex].

Check that you are correct by multiplying.

[latex]\left(3x+5\right)\left(3x - 5\right)=9x^2-15x+15x-25=9x^2-25[/latex]

[latex]9{x}^{2}-25=\left(3x+5\right)\left(3x - 5\right)[/latex]

The most helpful thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.

Factor [latex]81{y}^{2}-144[/latex].

Notice that [latex]81{y}^{2}[/latex] and [latex]144[/latex] are perfect squares because [latex]81{y}^{2}={\left(9x\right)}^{2}[/latex] and [latex]144={12}^{2}[/latex].

This means that [latex]a=9x,\text{ and }b=12[/latex]

The polynomial represents a difference of squares and can be rewritten as [latex]\left(9x+12\right)\left(9x - 12\right)[/latex].

[latex]\left(9x+12\right)\left(9x - 12\right)=81x^2-108x+108x-144=81x^2-144[/latex]

[latex]81{y}^{2}-144=\left(9x+12\right)\left(9x - 12\right)[/latex]

In the following video we show another example of how to use the formula for fact a difference of squares.

We can summarize the process for factoring a difference of squares with the shortcut this way:

How To: Given a difference of squares, factor it into binomials.

• Write the factored form as [latex]\left(a+b\right)\left(a-b\right)[/latex].

Think About It

Is there a formula to factor the sum of squares, [latex]a^2+b^2[/latex], into a product of two binomials?

Write down some ideas for how you would answer this in the box below before you look at the answer.

There is no way to factor a sum of squares into a product of two binomials, this is because of addition – the middle term needs to “disappear” and the only way to do that is with opposite signs.  to get a positive result, you must multiply two numbers with the same signs.

The only time a sum of squares can be factored is if they share any common factors, as in the following case:

[latex]9x^2+36[/latex]

[latex]9x^2={(3x)}^2, \text{ and }36 = 6^2[/latex]

The only way to factor this expression is by pulling out the GCF which is 9.

[latex]9x^2+36=9(x^2+4)[/latex]

(7.4.2)- Factoring Special Cases – Cubes

Some interesting patterns arise when you are working with cubed quantities within polynomials. Specifically, there are two more special cases to consider: [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}-b^{3}[/latex].

Let’s take a look at how to factor sums and differences of cubes.

Factoring the sum of cubes

The term “cubed” is used to describe a number raised to the third power. In geometry, a cube is a six-sided shape with equal width, length, and height; since all these measures are equal, the volume of a cube with width x can be represented by [latex]x^{3}[/latex]. (Notice the exponent!)

Cubed numbers get large very quickly. [latex]1^{3}=1[/latex], [latex]2^{3}=8[/latex], [latex]3^{3}=27[/latex], [latex]4^{3}=64[/latex], and [latex]5^{3}=125[/latex].

Before looking at factoring a sum of two cubes, let’s look at the possible factors.

It turns out that [latex]a^{3}+b^{3}[/latex] can actually be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]. Let’s check these factors by multiplying.

Does [latex](a+b)(a^{2}–ab+b^{2})=a^{3}+b^{3}[/latex]?

Apply the distributive property.

[latex]\left(a\right)\left(a^{2}–ab+b^{2}\right)+\left(b\right)\left(a^{2}–ab+b^{2}\right)[/latex]

Multiply by a .

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(b\right)\left(a^{2}-ab+b^{2}\right)[/latex]

Multiply by b .

[latex]\left(a^{3}–a^{2}b+ab^{2}\right)+\left(a^{2}b–ab^{2}+b^{3}\right)[/latex]

Rearrange terms in order to combine the like terms.

[latex]a^{3}-a^{2}b+a^{2}b+ab^{2}-ab^{2}+b^{3}[/latex]

Did you see that? Four of the terms cancelled out, leaving us with the (seemingly) simple binomial [latex]a^{3}+b^{3}[/latex]. So, the factors are correct.

You can use this pattern to factor binomials in the form [latex]a^{3}+b^{3}[/latex], otherwise known as “the sum of cubes.”

The Sum of Cubes

A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex].

The factored form of [latex]x^{3}+64[/latex] is [latex]\left(x+4\right)\left(x^{2}–4x+16\right)[/latex].

The factored form of [latex]8x^{3}+y^{3}[/latex] is [latex]\left(2x+y\right)\left(4x^{2}–2xy+y^{2}\right)[/latex].

Factor [latex]x^{3}+8y^{3}[/latex].

Identify that this binomial fits the sum of cubes pattern: [latex]a^{3}+b^{3}[/latex].

[latex]a=x[/latex], and [latex]b=2y[/latex] (since [latex]2y\cdot2y\cdot2y=8y^{3}[/latex]).

[latex]x^{3}+8y^{3}[/latex]

Factor the binomial as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=2y[/latex] into the expression.

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+\left(2y\right)^{2}\right)[/latex]

Square [latex](2y)^{2}=4y^{2}[/latex].

[latex]\left(x+2y\right)\left(x^{2}-x\left(2y\right)+4y^{2}\right)[/latex]

Multiply [latex]−x\left(2y\right)=−2xy[/latex] (writing the coefficient first).

And that’s it. The binomial [latex]x^{3}+8y^{3}[/latex] can be factored as [latex]\left(x+2y\right)\left(x^{2}–2xy+4y^{2}\right)[/latex]! Let’s try another one.

You should always look for a common factor before you follow any of the patterns for factoring.

Factor [latex]16m^{3}+54n^{3}[/latex].

Factor out the common factor 2.

[latex]16m^{3}+54n^{3}[/latex]

[latex]8m^{3}[/latex] and [latex]27n^{3}[/latex] are cubes, so you can factor [latex]8m^{3}+27n^{3}[/latex] as the sum of two cubes: [latex]a=2m[/latex], and [latex]b=3n[/latex].

[latex]2\left(8m^{3}+27n^{3}\right)[/latex]

Factor the binomial [latex]8m^{3}+27n^{3}[/latex] substituting [latex]a=2m[/latex] and [latex]b=3n[/latex] into the expression [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex].

[latex]2\left(2m+3n\right)\left[\left(2m\right)^{2}-\left(2m\right)\left(3n\right)+\left(3n\right)^{2}\right][/latex]

Square: [latex](2m)^{2}=4m^{2}[/latex] and [latex](3n)^{2}=9n^{2}[/latex].

[latex]2\left(2m+3n\right)\left[4m^{2}-\left(2m\right)\left(3n\right)+9n^{2}\right][/latex]

Multiply [latex]-\left(2m\right)\left(3n\right)=-6mn[/latex].

Having seen how binomials in the form [latex]a^{3}+b^{3}[/latex] can be factored, it should not come as a surprise that binomials in the form [latex]a^{3}-b^{3}[/latex] can be factored in a similar way.

The Difference of Cubes

A binomial in the form [latex]a^{3}–b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex].

The factored form of [latex]x^{3}–64[/latex] is [latex]\left(x–4\right)\left(x^{2}+4x+16\right)[/latex].

The factored form of [latex]27x^{3}–8y^{3}[/latex] is [latex]\left(3x–2y\left)\right(9x^{2}+6xy+4y^{2}\right)[/latex].

Notice that the basic construction of the factorization is the same as it is for the sum of cubes; the difference is in the [latex]+[/latex] and [latex]–[/latex] signs. Take a moment to compare the factored form of [latex]a^{3}+b^{3}[/latex] with the factored form of [latex]a^{3}-b^{3}[/latex].

Factored form of [latex]a^{3}+b^{3}[/latex]: [latex]\left(a+b\right)\left(a^{2}-ab+b^{2}\right)[/latex]

Factored form of [latex]a^{3}-b^{3}[/latex]: [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex]

This can be tricky to remember because of the different signs—the factored form of [latex]a^{3}+b^{3}[/latex] contains a negative, and the factored form of [latex]a^{3}-b^{3}[/latex] contains a positive! Some people remember the different forms like this:

“Remember one sequence of variables: [latex]a^{3}b^{3}=\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]. There are 4 missing signs. Whatever the first sign is, it is also the second sign. The third sign is the opposite, and the fourth sign is always [latex]+[/latex].”

Try this for yourself. If the first sign is [latex]+[/latex], as in [latex]a^{3}+b^{3}[/latex], according to this strategy how do you fill in the rest: [latex]\left(a\,b\right)\left(a^{2}ab\,b^{2}\right)[/latex]? Does this method help you remember the factored form of [latex]a^{3}+b^{3}[/latex] and [latex]a^{3}–b^{3}[/latex]?

Let’s go ahead and look at a couple of examples. Remember to factor out all common factors first.

Factor [latex]8x^{3}–1,000[/latex].

Factor out 8.

[latex]8(x^{3}–125)[/latex]

Identify that the binomial fits the pattern [latex]a^{3}-b^{3}:a=x[/latex], and [latex]b=5[/latex] (since [latex]5^{3}=125[/latex]).

[latex]8\left(x^{3}–125\right)[/latex]

Factor [latex]x^{3}–125[/latex] as [latex]\left(a–b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=x[/latex] and [latex]b=5[/latex] into the expression.

[latex]8\left(x-5\right)\left[x^{2}+\left(x\right)\left(5\right)+5^{2}\right][/latex]

Square the first and last terms, and rewrite [latex]\left(x\right)\left(5\right)[/latex] as [latex]5x[/latex].

[latex]8\left(x–5\right)\left(x^{2}+5x+25\right)[/latex]

Let’s see what happens if you don’t factor out the common factor first. In this example, it can still be factored as the difference of two cubes. However, the factored form still has common factors, which need to be factored out.

As you can see, this last example still worked, but required a couple of extra steps. It is always a good idea to factor out all common factors first. In some cases, the only efficient way to factor the binomial is to factor out the common factors first.

Here is one more example. Note that [latex]r^{9}=\left(r^{3}\right)^{3}[/latex] and that [latex]8s^{6}=\left(2s^{2}\right)^{3}[/latex].

Factor [latex]r^{9}-8s^{6}[/latex].

[latex]r^{9}-8s^{6}[/latex]

Rewrite [latex]r^{9}[/latex] as [latex]\left(r^{3}\right)^{3}[/latex] and rewrite [latex]8s^{6}[/latex] as [latex]\left(2s^{2}\right)^{3}[/latex].

[latex]\left(r^{3}\right)^{3}-\left(2s^{2}\right)^{3}[/latex]

Now the binomial is written in terms of cubed quantities. Thinking of [latex]a^{3}-b^{3}[/latex], [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex].

Factor the binomial as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex], substituting [latex]a=r^{3}[/latex] and [latex]b=2s^{2}[/latex] into the expression.

[latex]\left(r^{3}-2s^{2}\right)\left[\left(r^{3}\right)^{2}+\left(r^{3}\right)\left(2s^{2}\right)+\left(2s^{2}\right)^{2}\right][/latex]

Multiply and square the terms.

[latex]\left(r^{3}-2s^{2}\right)\left(r^{6}+2r^{3}s^{2}+4s^{4}\right)[/latex]

In the following two video examples we show more binomials that can be factored as a sum or difference of cubes.

You encounter some interesting patterns when factoring. Two special cases—the sum of cubes and the difference of cubes—can help you factor some binomials that have a degree of three (or higher, in some cases). The special cases are:

• A binomial in the form [latex]a^{3}+b^{3}[/latex] can be factored as [latex]\left(a+b\right)\left(a^{2}–ab+b^{2}\right)[/latex]
• A binomial in the form [latex]a^{3}-b^{3}[/latex] can be factored as [latex]\left(a-b\right)\left(a^{2}+ab+b^{2}\right)[/latex]

Always remember to factor out any common factors first.

(7.4.3) – More factoring methods

Factoring using substitution.

We are going to move back to factoring polynomials – our exponents will be positive integers. Sometimes we encounter a polynomial that looks similar to something we know how to factor, but isn’t quite the same. Substitution is a useful tool that can be used to “mask” a term or expression to make algebraic operations easier.

You may recall that substitution can be used to solve systems of linear equations, and to check whether a point is a solution to a system of linear equations. for example:

To determine whether the ordered pair [latex]\left(5,1\right)[/latex] is a solution to the given system of equations.

We can substitute the ordered pair [latex]\left(5,1\right)[/latex] into both equations.

[latex]\begin{array}{ll}\left(5\right)+3\left(1\right)=8\hfill & \hfill \\ \text{ }8=8\hfill & \text{True}\hfill \\ 2\left(5\right)-9=\left(1\right)\hfill & \hfill \\ \text{ }\text{1=1}\hfill & \text{True}\hfill \end{array}[/latex]

We replaced the variable with a number and then performed the algebraic operations specified.  In the next example we will see how we can use a similar technique to factor a fourth degree polynomial.

Factor [latex]x^4+3x^2+2[/latex]

This looks a lot like a trinomial that we know how to factor – [latex]x^2+3x+2=(x+2)(x+1)[/latex] except for the exponents.

If we substitute [latex]u=x^2[/latex], and recognize that [latex]u^2=(x^2)^2=x^4[/latex] we may be able to factor this beast!

Everywhere there is an [latex]x^2[/latex] we will replace it with a u, then factor.

[latex]u^2+3u+2=(u+1)(u+2)[/latex]

We aren’t quite done yet, we want to factor the original polynomial which had x as it’s variable, so we need to replace [latex]x^2=u[/latex] now that we are done factoring.

[latex](u+1)(u+2)=(x^2+1)(x^2+2)[/latex]

[latex]x^4+3x^2+2=(x^2+1)(x^2+2)[/latex]

In the following video we show two more examples of how to use substitution to factor a fourth degree polynomial and an expression with fractional exponents.

Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example we will first find the GCF of a trinomial, and after factoring it out we will be able to factor again so that we end up with a product of a monomial, and two binomials.

Factor completely [latex]6m^2k-3mk-3k[/latex].

Factor 3k from the trinomial:

[latex]6m^2k-3mk-3k=3k\left(2m^2-m-1\right)[/latex]

We are left with a trinomial that can be factored using your choice of factoring method. We will create a table to find the factors of [latex]2\cdot{-1}=-2[/latex] that sum to [latex]-1[/latex]

Our factors are [latex]-2,1[/latex], so we can factor by grouping:

Rewrite the middle term with the factors we found with the table:

[latex]\left(2m^2-m-1\right)=2m^2-2m+m-1[/latex]

Regroup and find the GCF of each group:

[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[/latex]

Now factor [latex](m-1)[/latex] from each term:

[latex]2m^2-m-1=(m-1)(2m+1)[/latex]

Don’t forget the original GCF that we factored out! Our final factored form is:

[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[/latex]

In our last example we shoe that it is important to factor out a GCF if there is one before you being using the techniques shown in this module.

(7.4.4) – General Factoring Strategy

With so many different tools used to factor, it is easy to get lost as to which tool to use when. Here we will attempt to organize all the different factoring types we have seen. A large part of deciding how to solve a problem is based on how many terms are in the problem. For all problem types we will always try to factor out the GCF first.

Factoring Strategy

1) Factor out the GCF! (lookout for substitutions!)

2) 2 terms : sum or difference of squares or cubes:

• [latex]a^2 - b^2 = (a + b) (a - b)[/latex]
• [latex]a^2 + b^2 = \text{ Prime}[/latex]
• [latex]a^3 + b^3 = (a + b) (a^2 - a b + b^2)[/latex]
• [latex]a^3 - b^3 = (a - b) (a^2 + \text{ ab} + b^2)[/latex]

3) 3 terms: watch for perfect square!

• [latex]a^2 + 2 a b + b^2 = (a + b)^2[/latex]
• [latex]a^2 - 2ab +b^2 = (a-b)^2[/latex]
• factoring by trial and error

4) 4 terms :

• Screenshot: Method to the Madness. Provided by : Lumen Learning. License : CC BY: Attribution
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2.3.1: Systems of Linear Equations – Special Cases (Exercises)

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• Rupinder Sekhon and Roberta Bloom
• De Anza College

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SECTION 2.3 PROBLEM SET: SYSTEMS OF LINEAR EQUATIONS - SPECIAL CASES

Solve the following inconsistent or dependent systems by using the Gauss-Jordan method.