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• factor\:x^{2}-5x+6
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• What is the sum of cubes formula?
• The sum of cubes formula is a³ + b³ = (a+b)(a² - ab + b²)
• What is the difference of squares formula?
• The difference of squares formula is a² - b² = (a+b)(a-b)
• What is the difference of cubes formula?
• The difference of cubes formula is a³ - b³ = (a-b)(a² + ab + b²)
• How do you solve factoring by greatest common monomial factor?
• To factor by greatest common monomial factor, find the greatest common monomial factor among the terms of the expression and then factor it out of each term.
• How do you factor a monomial?
• To factor a monomial, write it as the product of its factors and then divide each term by any common factors to obtain the fully-factored form.
• How do you factor a binomial?
• To factor a binomial, write it as the sum or difference of two squares or as the difference of two cubes.
• How do you factor a trinomial?
• To factor a trinomial x^2+bx+c find two numbers u, v that multiply to give c and add to b. Rewrite the trinomial as the product of two binomials (x-u)(x-v)
• How to find LCM with the listing multiples method?
• To find the LCM of two numbers using the listing multiples method write down the multiples of the first number and write down the multiples of the second number. Find the smallest number that is a multiple of both of the numbers.

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• Middle School Math Solutions – Polynomials Calculator, Factoring Quadratics Just like numbers have factors (2×3=6), expressions have factors ((x+2)(x+3)=x^2+5x+6). Factoring is the process...

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The process of factoring is essential to the simplification of many algebraic expressions and is a useful tool in solving higher degree equations. In fact, the process of factoring is so important that very little of algebra beyond this point can be accomplished without understanding it.

In earlier chapters the distinction between terms and factors has been stressed. You should remember that terms are added or subtracted and factors are multiplied. Three important definitions follow.

Terms occur in an indicated sum or difference. Factors occur in an indicated product.

• An expression is in factored form only if the entire expression is an indicated product.

Note in these examples that we must always regard the entire expression. Factors can be made up of terms and terms can contain factors, but factored form must conform to the definition above.

Factoring is a process of changing an expression from a sum or difference of terms to a product of factors.

Note that in this definition it is implied that the value of the expression is not changed - only its form.

REMOVING COMMON FACTORS

• Determine which factors are common to all terms in an expression.
• Factor common factors.

In the previous chapter we multiplied an expression such as 5(2x + 1) to obtain 10x + 5. In general, factoring will "undo" multiplication. Each term of 10x + 5 has 5 as a factor, and 10x + 5 = 5(2x + 1).

To factor an expression by removing common factors proceed as in example 1.

Next look for factors that are common to all terms, and search out the greatest of these. This is the greatest common factor. In this case, the greatest common factor is 3x.

Proceed by placing 3x before a set of parentheses.

The terms within the parentheses are found by dividing each term of the original expression by 3x.

If we had only removed the factor "3" from 3x 2 + 6xy + 9xy 2 , the answer would be

3(x 2 + 2xy + 3xy 2 ).

Multiplying to check, we find the answer is actually equal to the original expression. However, the factor x is still present in all terms. Hence, the expression is not completely factored.

• It must be possible to multiply the factored expression and get the original expression.
• FThe expression must be completely factored.

Example 2 Factor 12x 3 + 6x 2 + 18x.

At this point it should not be necessary to list the factors of each term. You should be able to mentally determine the greatest common factor. A good procedure to follow is to think of the elements individually. In other words, don�t attempt to obtain all common factors at once but get first the number, then each letter involved. For instance, 6 is a factor of 12, 6, and 18, and x is a factor of each term. Hence 12x 3 + 6x 2 + 18x = 6x(2x 2 + x + 3). Multiplying, we get the original and can see that the terms within the parentheses have no other common factor, so we know the solution is correct.

If an expression cannot be factored it is said to be prime .

FACTORING BY GROUPING

• Factor expressions when the common factor involves more than one term.
• Factor by grouping.

An extension of the ideas presented in the previous section applies to a method of factoring called grouping .

First we must note that a common factor does not need to be a single term. For instance, in the expression 2y(x + 3) + 5(x + 3) we have two terms. They are 2y(x + 3) and 5(x + 3). In each of these terms we have a factor (x + 3) that is made up of terms. This factor (x + 3) is a common factor.

Sometimes when there are four or more terms, we must insert an intermediate step or two in order to factor.

First note that not all four terms in the expression have a common factor, but that some of them do. For instance, we can factor 3 from the first two terms, giving 3(ax + 2y). If we factor a from the remaining two terms, we get a(ax + 2y). The expression is now 3(ax + 2y) + a(ax + 2y), and we have a common factor of (ax + 2y) and can factor as (ax + 2y)(3 + a). Multiplying (ax + 2y)(3 + a), we get the original expression 3ax + 6y + a 2 x + 2ay and see that the factoring is correct.

This is an example of factoring by grouping since we "grouped" the terms two at a time.

Sometimes the terms must first be rearranged before factoring by grouping can be accomplished.

Example 7 Factor 3ax + 2y + 3ay + 2x.

The first two terms have no common factor, but the first and third terms do, so we will rearrange the terms to place the third term after the first. Always look ahead to see the order in which the terms could be arranged.

In all cases it is important to be sure that the factors within parentheses are exactly alike. This may require factoring a negative number or letter.

Example 8 Factor ax - ay - 2x + 2y.

Note that when we factor a from the first two terms, we get a(x - y). Looking at the last two terms, we see that factoring +2 would give 2(-x + y) but factoring "-2" gives - 2(x - y). We want the terms within parentheses to be (x - y), so we proceed in this manner.

FACTORING TRINOMIALS

• Mentally multiply two binomials.
• Factor a trinomial having a first term coefficient of 1.
• Find the factors of any factorable trinomial.

A large number of future problems will involve factoring trinomials as products of two binomials. In the previous chapter you learned how to multiply polynomials. We now wish to look at the special case of multiplying two binomials and develop a pattern for this type of multiplication.

Since this type of multiplication is so common, it is helpful to be able to find the answer without going through so many steps. Let us look at a pattern for this.

From the example (2x + 3)(3x - 4) = 6x 2 + x - 12, note that the first term of the answer (6x 2 ) came from the product of the two first terms of the factors, that is (2x)(3x).

Also note that the third term (-12) came from the product of the second terms of the factors, that is ( + 3)(-4).

We now have the following part of the pattern:

Now looking at the example again, we see that the middle term (+x) came from a sum of two products (2x)( -4) and (3)(3x).

• First term by first term
• Outside terms
• Inside terms
• Last term by last term

These products are shown by this pattern.

When the products of the outside terms and inside terms give like terms, they can be combined and the solution is a trinomial.

You should memorize this pattern.

Not only should this pattern be memorized, but the student should also learn to go from problem to answer without any written steps. This mental process of multiplying is necessary if proficiency in factoring is to be attained.

As you work the following exercises, attempt to arrive at a correct answer without writing anything except the answer. The more you practice this process, the better you will be at factoring.

Now that we have established the pattern of multiplying two binomials, we are ready to factor trinomials. We will first look at factoring only those trinomials with a first term coefficient of 1.

Since this is a trinomial and has no common factor we will use the multiplication pattern to factor.

First write parentheses under the problem.

We now wish to fill in the terms so that the pattern will give the original trinomial when we multiply. The first term is easy since we know that (x)(x) = x 2 .

We must now find numbers that multiply to give 24 and at the same time add to give the middle term. Notice that in each of the following we will have the correct first and last term.

Only the last product has a middle term of 11x, and the correct solution is

This method of factoring is called trial and error - for obvious reasons.

Here the problem is only slightly different. We must find numbers that multiply to give 24 and at the same time add to give - 11. You should always keep the pattern in mind. The last term is obtained strictly by multiplying, but the middle term comes finally from a sum. Knowing that the product of two negative numbers is positive, but the sum of two negative numbers is negative, we obtain

We are here faced with a negative number for the third term, and this makes the task slightly more difficult. Since -24 can only be the product of a positive number and a negative number, and since the middle term must come from the sum of these numbers, we must think in terms of a difference. We must find numbers whose product is 24 and that differ by 5. Furthermore, the larger number must be negative, because when we add a positive and negative number the answer will have the sign of the larger. Keeping all of this in mind, we obtain

• When the sign of the third term is positive, both signs in the factors must be alike-and they must be like the sign of the middle term.
• When the sign of the last term is negative, the signs in the factors must be unlike-and the sign of the larger must be like the sign of the middle term.

In the previous exercise the coefficient of each of the first terms was 1. When the coefficient of the first term is not 1, the problem of factoring is much more complicated because the number of possibilities is greatly increased.

Notice that there are twelve ways to obtain the first and last terms, but only one has 17x as a middle term.

There is only one way to obtain all three terms:

In this example one out of twelve possibilities is correct. Thus trial and error can be very time-consuming.

Even though the method used is one of guessing, it should be "educated guessing" in which we apply all of our knowledge about numbers and exercise a great deal of mental arithmetic. In the preceding example we would immediately dismiss many of the combinations. Since we are searching for 17x as a middle term, we would not attempt those possibilities that multiply 6 by 6, or 3 by 12, or 6 by 12, and so on, as those products will be larger than 17. Also, since 17 is odd, we know it is the sum of an even number and an odd number. All of these things help reduce the number of possibilities to try.

• The last term is positive, so two like signs.
• The middle term is negative, so both signs will be negative.
• The factors of 6x2 are x, 2x, 3x, 6x. The factors of 15 are 1, 3, 5, 15.
• Eliminate as too large the product of 15 with 2x, 3x, or 6x. Try some reasonable combinations.

• The last term is negative, so unlike signs.
• We must find products that differ by 5 with the larger number negative.
• We eliminate a product of 4x and 6 as probably too large.
• Try some combinations.

(4x - 3)(x + 2) : Here the middle term is + 5x, which is the right number but the wrong sign. Be careful not to accept this as the solution, but switch signs so the larger product agrees in sign with the middle term.

SPECIAL CASES IN FACTORING

• Identify and factor the differences of two perfect squares.
• Identify and factor a perfect square trinomial.

In this section we wish to examine some special cases of factoring that occur often in problems. If these special cases are recognized, the factoring is then greatly simplified.

The first special case we will discuss is the difference of two perfect squares .

Recall that in multiplying two binomials by the pattern, the middle term comes from the sum of two products.

From our experience with numbers we know that the sum of two numbers is zero only if the two numbers are negatives of each other.

In each example the middle term is zero. Note that if two binomials multiply to give a binomial (middle term missing), they must be in the form of (a - b) (a + b).

Here both terms are perfect squares and they are separated by a negative sign.

Special cases do make factoring easier, but be certain to recognize that a special case is just that-very special. In this case both terms must be perfect squares and the sign must be negative, hence "the difference of two perfect squares."

You must also be careful to recognize perfect squares. Remember that perfect square numbers are numbers that have square roots that are integers. Also, perfect square exponents are even.

Another special case in factoring is the perfect square trinomial. Observe that squaring a binomial gives rise to this case.

• The first term is a perfect square.
• The third term is a perfect square.
• The middle term is twice the product of the square root of the first and third terms.
• 25x 2 is a perfect square-principal square root = 5x.
• 4 is a perfect square-principal square root = 2.
• 20x is twice the product of the square roots of 25x 2 and
• 20x = 2(5x)(2).

To factor a perfect square trinomial form a binomial with the square root of the first term, the square root of the last term, and the sign of the middle term, and indicate the square of this binomial.

Thus, 25x 2 + 20x + 4 = (5x + 2) 2

Not the special case of a perfect square trinomial.

OPTIONAL SHORTCUTS TO TRIAL AND ERROR FACTORING

• Find the key number of a trinomial.
• Use the key number to factor a trinomial.

In this section we wish to discuss some shortcuts to trial and error factoring. These are optional for two reasons. First, some might prefer to skip these techniques and simply use the trial and error method; second, these shortcuts are not always practical for large numbers. However, they will increase speed and accuracy for those who master them.

The first step in these shortcuts is finding the key number . After you have found the key number it can be used in more than one way.

In a trinomial to be factored the key number is the product of the coefficients of the first and third terms.

The first use of the key number is shown in example 3.

A second use for the key number as a shortcut involves factoring by grouping. It works as in example 5.

COMPLETE FACTORIZATION

• First look for common factors.
• Factor the remaining trinomial by applying the methods of this chapter.

We have now studied all of the usual methods of factoring found in elementary algebra. However, you must be aware that a single problem can require more than one of these methods. Remember that there are two checks for correct factoring.

• Will the factors multiply to give the original problem?
• Are all factors prime?

A good procedure to follow in factoring is to always remove the greatest common factor first and then factor what remains, if possible.

• Factoring is a process that changes a sum or difference of terms to a product of factors.
• A prime expression cannot be factored.
• The greatest common factor is the greatest factor common to all terms.
• An expression is completely factored when no further factoring is possible.
• The possibility of factoring by grouping exists when an expression contains four or more terms.
• The FOIL method can be used to multiply two binomials.
• Special cases in factoring include the difference of two squares and perfect square trinomials .
• The key number is the product of the coefficients of the first and third terms of a trinomial.
• To remove common factors find the greatest common factor and divide each term by it.

• To factor a perfect square trinomial form a binomial with the square root of the first term, the square root of the last term, and the sign of the middle term and indicate the square of this binomial.
• Use the key number as an aid in determining factors whose sum is the coefficient of the middle term of a trinomial.

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6: Factoring and Solving by Factoring

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• 6.1: Introduction to Factoring
• 6.2: Factoring Trinomials of the Form x²+bx+c
• 6.3: Factoring Trinomials of the Form ax²+bx+c
• 6.4: Factoring Special Binomials
• 6.5: General Guidelines for Factoring Polynomials
• 6.6: Solving Equations by Factoring
• 6.7: Applications Involving Quadratic Equations
• 6.E: Review Exercises and Sample Exam

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Factorising

Here is everything you need to know about factorising for GCSE maths (Edexcel, AQA and OCR). You’ll learn the essentials of factorising expressions and factorising quadratics including factorising into single brackets and double brackets.

Look out for the factorising worksheets and exam questions at the end.

What is factorising

Factorising is the reverse process of expanding brackets. To factorise an expression fully, means to put it in brackets by taking out the highest common factors.

The simplest way of factorising is:

• Find the highest common factor of each of the terms in the expression.
• Write the highest common factor (HCF) in front of any brackets
• Fill in each term in the brackets by multiplying out.

However there are different ways to factorise different types of algebraic expressions; we will learn about them all here.

What is factorising?

Factorising (mixed) worksheet

Get your free factorising worksheet of 20+ questions and answers. Includes reasoning and applied questions.

How to factorise expressions

To factorise algebraic expressions there are three basic methods. When you are factorising quadratics you will usually use the double brackets or difference of two squares method.

1. Factorising single brackets

Example of factorising an algebraic expression:

Remember: 3x+6 is known as a binomial because it is an expression with two terms

2. Factorising double brackets

a) When factorising quadratic expressions in the form x 2 + b x + c

b) When factorising quadratic expressions in the form a x 2 + b x + c

Remember: Expressions with three terms like x 2 + 6 x + 5 and 2 x 2 + 5 x + 3 are known as a trinomials.

3. Differences of two squares

Using the difference of two squares:

Explain how to factorise expressions

Factorisation methods

Each method of factorising or factoring expressions is summarised below. For detailed examples, practice questions and worksheets on each one follow the links to the step by step guides.

Factorising example using single brackets

To factorise fully:

• Find the highest common factor (HCF) of the numbers 3 (the coefficient of x ) and 6 (the constant).

Factors of 3 : 1, 3

Factors of 6 : 1, 6 2, 3

Top tip: Writing the factor pairs makes it easier to list all the factors

The highest common factor (HCF) of 3 x and 6 is 3

2 Write the highest common factor (HCF) at the front of the single bracket.

3 Fill in each term in the bracket by multiplying out.

What do I need to multiply 3 by to give me 3 x ?

What do I need to multiply 3 by to give me 6 ?

We can check the answer by multiplying out the bracket!

Step by step guide: Factorising single brackets

2a) Factorising quadratics into double brackets: x 2 + bx + c

Factorising example for quadratic expressions in the form x 2 + b x + c

• Write out the factor pairs of the last number ( 5 )

Factors of 5 : 1 , 5

2 Find a pair of factors that + to give the middle number ( 6 ) and ✕ to give the last number ( 5 ).

3 Write two brackets and put the variable at the start of each one.

4 Write one factor in the first bracket and the other factor in the second bracket. The order isn’t important, the signs of the factors are.

2b) Factorising quadratics into double brackets: ax 2 + bx + c

Factorising example for quadratic expressions in the form a x 2 + b x + c

• Multiply the end numbers together ( 2 and 3 ) then write out the factor pairs of this new number in order

Factors of 6: 1, 6 2, 3

2 We need a pair of factors that + to give the middle number ( 5 ) and ✕ to give this new number (6)

2 + 3 = 5 ✔

2 ✕ 3 = 6 ✔

3 Rewrite the original expression, this time splitting the middle term into the two factors we found in step 2.

4 Split the equation down the middle and factorise fully each half.

5 Factorise the whole expression by bringing whatever is in the bracket to the front and writing the two other terms in the other bracket.

Step by step guide: Factorising quadratics

3. Difference of two squares

Factorising example using difference of two squares:

• Write down 2 brackets.

2 Square root the first term and write it on the left-hand side of both brackets.

3 Square root the last term and write it on the right-hand side of both brackets.

4 Put + in the middle of one bracket and – in the middle of the other (the order doesn’t matter).

Step by step guide: Difference of two squares

Practice factorising questions (mixed)

1. Fully factorise:

The highest common factor of 10 and 5 is 5 . Therefore, we can divide the original expression by 5 , which means the bracket must contain 2-y .

2. Fully factorise:

The highest common factor of 20x^{2} and 8x is 4x . Therefore, we can divide the original expression by 4x , which means the bracket must contain 5x-2 .

3. Fully factorise:

In this case, we are looking for numbers that multiply to make -6 and add to make -1 . By considering the factor pairs of -6 , we conclude that we can use +2 and -3 .

4. Fully factorise:

The original expression can be divided by 2 , and then written as a product of two brackets.

Divide each term by 2 , and rewrite.

To factorise the quadratic expression, we are looking for numbers that multiply to -3 and sum to -2 . By considering factor pairs, we conclude that we need to use +1 and -3 .

5. Fully factorise:

This is a special case (difference of two squares), which means we can take square roots of the coefficient of x and the constant term, then write one bracket with a + sign and the other bracket with a – sign.

6. Fully factorise:

Factorising GCSE questions (mixed)

1. Factorise: 9x − 18

2. Factorise fully: 16x 2 + 20xy

4x(4x + 5y)

3. Factorise fully: 3y 2 − 4y − 4

(3y + 2)(y − 2)

Learning checklist

You have now learnt how to:

• Manipulate algebraic expressions by taking out common factors to factorise into a single bracket.
• Factorise quadratic expressions of the form x 2 + bx + c
• Factorise quadratic expressions of the form of the difference of two squares.
• Factorise quadratic expressions of the form ax 2 + bx + c (H)

The next lessons are

• Solving equations
• Expanding brackets
• Rearranging equations

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Factorisation Problems

Factorisation problems are provided here with solutions for students. Factorisation questions and solutions for students of Class 7, Class 8, Class 9 and Class 10 are given to make them practise algebra and polynomial concepts. The factorisation is a method of factoring a number or a polynomial. The polynomials are decomposed into products of their factors. For example, the factorisation of x 2 + 2x is x(x + 2), where x and x+2 are the factors that can be multiplied together to get the original polynomial.

Now let’s solve some factorisation problems here to understand better. Also, get some practice questions based on the factorisation at the end of the article.

Factorisation Problems and Solutions

Q.1: Factorise 4x 2 + 12x + 5.

Solution: 4x 2 + 12x + 5

We can write the above expression as:

⇒ 4x 2 + 10x + 2x + 5

Taking the common terms out;

⇒ 2x(2x + 5) + 1 (2x + 5)

⇒ (2x + 1) (2x + 5)

Q.2: Factorise y 2 + 16y + 60.

Solution: y 2 + 16y + 60

⇒ y 2 + 10y + 6y + 60

⇒ y(y + 10) + 6 (y + 10)

⇒ (y + 6) (y + 10)

Q.3: Factorise 5x 2 + 14x – 3.

Solution: 5x 2 + 14x – 3

⇒ 5x 2 – x + 15x – 3

⇒ x(5x – 1) + 3(5x – 1)

⇒ (5x – 1) (x + 3)

Q.4: Factorise 4(x+y) 2 – 28(x 2 – y 2 ) + 49(x-y) 2 .

Solution: Given,

4(x+y) 2 – 28(x 2 – y 2 ) + 49(x-y) 2

Since, by the algebraic formula we know;

x 2 – y 2 = (x + y) (x – y)

⇒ 4(x+y) 2 – 28(x – y) (x + y) + 49(x-y) 2

⇒ [2(x+y) – 7(x-y)] 2

⇒ [2x + 2y – 7x – 7y] 2

⇒ (9y – 5x) 2

Q.5: Factorise 25(a+b) 2 – (a-b) 2 .

Solution: 25(a+b) 2 – (a-b) 2

We can write the above expression as;

By the algebraic formula, we know that, a 2 – b 2 = (a+b) (a-b).

Thus, we can write the expression as:

Q.6: Factor 6a 2 b − 8ab + 10ab 2 .

Solution: In the given expression, the highest common factor is 2ab, therefore taking 2ab as common, we get;

6a 2 b − 8ab + 10ab 2 = 2ab(3a − 4 + 5b)

Q.7: Solve (4x 2 – 25y 2 )/(2x + 5y).

Solution: Given, (4x 2 – 25y 2 )/(2x + 5y)

We can write, 4x 2 – 25y 2 as:

⇒ (2x) 2 – (5y) 2 = (2x + 5y) (2x – 5y)

⇒ [(2x + 5y) (2x – 5y)]/(2x + 5y)

⇒ 2x – 5y

Q.8: Factorise 2axy 2 + 10x + 3ay 2 + 15.

Solution: 2axy 2 + 10x + 3ay 2 + 15

Rearranging the terms we have;

⇒ (2axy 2 + 3ay 2 ) + (10x + 15)

Taking the common terms we get;

⇒ ay 2 (2x + 3) +5(2x + 3)

Hence, the required factors are:

⇒ (2x + 3) (ay 2 + 5)

Q.9: Factorise x 4 – (x – y) 4 .

Solution: Given, x 4 – (x – y) 4

It can be written as:

⇒ (x 2 ) 2 – [(x – y) 2 ] 2

Using algebraic formula, a 2 – b 2 = (a – b) (a + b)

Using (a-b) 2 = a 2 – 2ab + b 2 , identity we get;

⇒ (2x – y) y(2x 2 – 2xy + y 2 )

⇒ y(2x – y) (2x 2 – 2xy + y 2 )

Q.10: Simplify the following:

\(\begin{array}{l}\frac{(x-1)(x-2)\left(x^{2}-9 x+14\right)}{(x-7)\left(x^{2}-3 x+2\right)}\end{array} \)

Cancelling the common terms from numerator and denominator, we get;

= x – 2

Extra Questions for Factorisation

• Factor 18 x 3 +3 x 2 – 6 x .
• Factorise 81x 2 − 36 x + 4
• Find the factors of 7x 2 yz − 8y.
• Factor x 2 − 11x + 24
• Factorise 11x 2 + 33x − 110
• Factorise −16 + 49x 2
• What are the factors of the expression 72 – 2x 2 ?
• Find the factors of 9x 2 +4y 2 +12xy.
• Using the algebraic identities, factorise x 2 + 6x + 9.
• If one of the factors of (5x 2 + 70x – 160) is (x – 2), find the other factor.

Related Articles

• Factorisation Of Algebraic Expression
• Factorization of Polynomials
• Polynomials
• What is Prime Factorisation?
• Class 8 Maths Chapter 14 Factorisation MCQs
• Class 9 Maths Chapter 2 Polynomials MCQs
• Class 10 Maths Chapter 2 Polynomials MCQs

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Factoring to Solve Problems

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TEKS Standards and Student Expectations

A(10)  Number and algebraic methods. The student applies the mathematical process standards and algebraic methods to rewrite in equivalent forms and perform operations on polynomial expressions. The student is expected to:

A(10)(E) factor, if possible, trinomials with real factors in the form a x 2 + b x + c including perfect square trinomials of degree two

A(10)(F)  decide if a binomial can be written as the difference of two squares and if possible, use the structure of a difference of two squares to rewrite the binomial

A(10)(D) rewrite polynomial expressions of degree one and degree two in equivalent forms using the distributive property

A(8)  Quadratic functions and equations. The student applies the mathematical process standards to solve, with and without technology, quadratic equations and evaluate the reasonableness of their solutions. The student formulates statistical relationships and evaluates their reasonableness based on real-world data. The student is expected to:

A(8)(A) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula

Resource Objective(s)

The student will use a variety of methods to solve problems by factoring, including models, guess and check, grouping, and special cases.

Essential Questions

What is the process for factoring a trinomial with a leading coefficient of 1?

What is the process for factoring a trinomial when the leading coefficient is not 1?

How can you determine if the rules for difference of two squares, difference of two cubes, or perfect squares  can be used to factor?

When does the grouping method for factoring apply?

• GCF (Greatest Common Factor)
• Leading Coefficient
• Quadratic Equation

Introduction: Factoring by Modeling

One way to find the factors of a value is to use models to create a rectangle. The length and width represent the factors.

Example 1 : Factors of 12 can be shown three different ways. 

12 has a total of 6 different factors: 1, 2, 3, 4, 6, and 12.

Example 2:  Algebraic expressions can also be factored using algebra tiles to create rectangles. In the example below, 2 x + 4 is used to create a rectangle. 

As you can see, the length and width of the rectangle represent the factors of 2 x + 4. The length is ( x + 2) and the width is 2. 

The factors of 2 x + 4 are 2 and ( x + 2).

Example 3:  Trinomials can also be factored using algebra tiles. The trinomial x 2 + x - 6 can be modeled by the following rectangle.

The side lengths of the rectangle show the factors of x 2  + x - 6  are ( x + 3) and ( x - 2) .

When you are ready, complete the two practice examples below to check your understanding.

Factoring Rule 1: Greatest Common Factor (GCF)

Always check to see if you can factor something out!

The first rule to factoring is to find the greatest common factor (GCF) of each term in the polynomial .

• If there is any factor in common in the polynomial, divide each term by that factor.
• Then, rewrite the polynomial using the distributive property with the common factor on the outside of the parenthesis.

2 x 3 -   2 x 2 - 12 x

All terms in the example have a coefficient that is divisible by 2. Each term also has an x . Therefore, the GCF is 2 x .

When 2 x  is factored out and the distributive property is applied the result is 

2 x ( x 2 -   x   -   6 ) .

Check your understanding with the two practice examples  below.

Factoring Rule 2: Special Products

If you are given a binomial to factor, the first special product to check for is the difference of two squares.

Difference of Two Squares:

  a 2 - b 2 =   ( a   -   b ) ( a   +   b ) .

If both terms are separated by a subtraction symbol and they are both perfect squares, the square roots can be used to write the factors.

Example 1: Factor  4 x 2 - 9 y 2 :

• The terms are separated by a subtraction symbol.
• Both terms are perfect squares.

    ( 2 x ) 2   a n d   ( 3 y ) 2

According to the rule for the difference of two squares, the factors will be  ( 2 x - 3 y ) ( 2 x + 3 y ) .

If you are given a trinomial to factor, the first special product to check for is a perfect square trinomial.

Perfect Square Trinomials:

a 2 + 2 a b + b 2   =   ( a + b ) 2   o r   a 2 - 2 a b + b 2   =   ( a - b ) 2

If the first term is a square, the last term is a square, and the middle term is two times the square root of the first and the last term, then it is a perfect square. There are two forms: one with a positive middle term and one with a negative middle term.

Example 2: Factor  x 2 + 6 x + 9

Notice the first and last terms are both perfect squares. When you multiply the square roots of x  and 3 times 2, you get 6 x   which is the middle term. According to the rule, t he factors are 

( x + 3 ) ( x + 3 )   o r   ( x + 3 ) 2

Example 3: Factor  4 x 2 - 20 x + 25

Notice the first and last terms are both perfect squares. When you multiply the square roots of 2 x and 5 times -2, you get -20 x which is the middle term. According to the rule, the factors are 

( 2 x - 5 ) ( 2 x - 5 )   o r   ( 2 x - 5 ) 2

Check your understanding by completing the practice example below. Determine which polynomials can be factored with a special product rule. Match each polynomial with the type of special product (first blank) and its factors (second blank).

Factoring Rule 3: Trinomials with a Leading Coefficient of 1

After checking for a GCF and Special Products, look for polynomials with three terms that have a leading coefficient of 1. If the leading coefficient is 1, put x 's as the first term in each set of parentheses. Then you factor the last term and find the two factors that add or subtract to make the middle term.

Use the practice example to complete the activity below.

Factoring Rule 4: Other Trinomials

If the leading coefficient is not 1, you factor both the first and last terms and find the two products that add or subtract to make the middle term. Check your answer by using the smiley face check.

Factoring Rule 5: Four or More Terms

If there are four or more terms, you need to group them together, and then factor each part.

You can do this two ways—either by using the box method or parentheses.

Box Method: 

• Put all four terms in a two by two box.
• Look for the GCF in each column and row.
• Include signs for each of the factors.
• The result will be two factors: one on the length and the other on the width.

Parentheses Method: 

• Put parentheses around the first two terms and another set around the last two terms.
• Find the GCF of each set of parentheses. 
• Once the GCF is factored out, the terms inside each set of parentheses should match. 
• The factors will be the terms inside the parentheses (from step 3) and the GCF outside each set of parentheses (in step 2). 

Check your understanding by factoring the following polynomial (using either method) and answer the questions below. 

Solving Problems by Factoring

Some problems may ask for the solution of a quadratic equation . To solve these problems, set the factors equal to 0 and solve for x . 

     2 x 3   -    2 x 2   - 12 x   =   2 x   ( x 2   -    x   -    6 )   =   2 x   ( x   -    3 ) ( x   +   2 )

To solve this problem, you would need to set all factors to 0 and solve for x . There will be three solutions.

Fill in the blanks below to find the three solutions to the polynomial. 

Check polynomials in your graphing calculator:

Your solutions will be the x -intercepts on the graph.

• Input your equation using the equation editor [ Y  =] key on the graphing calculator.
• Press [GRAPH].
• Press [2nd] [TRACE] to access the CALC menu.
• Press [2] to select the zero command.
• Use the arrow keys to select a left bound and a right bound.

Vocabulary Activity

Journal activity.

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