# Factoring Polynomials: 5 Methods & Examples

Previous topic about polynomial is multiplying. If there are two or more simple polynomial that are multiplied it will become complex polynomial.

Concept of factoring polynomials is inversely. Starting from complex polynomial become simple polynomial.

If the polynomial is divided by one of the simple polynomials then the remainder is zero. It is also called as factor theorem.

If there is polynomial p(x) that has degree greater than or equal to one, then:

1. (x-a) is a factr of p(x), if p(a) = 0
2. If p(a) = 0 then (x-a) is a factor of p(x).

There are some factoring polynomial’s methods:

## Greatest Common Factor (GCF)

If there is ax+bx = x(a+b) then 2x4+4x2 = 2x2(x2+2).

It is because 2x2 is the GCF of 2x4+4x2.

It is also can be written as 2x4+4x2 = (2x2)(x2)+(2x2)(2) = 2x2(x2+2).

In another words, there are some steps using GCF method.

1. Find the GCF of all of terms in polynomial.
2. Write / express each term as multiplying of GCF and the another.
3. Take out the GCF from each term.

Knowing it is true or false factoring polynomial’s result, it can be checked by multiplying the result. It must get the complex polynomial (before it be factorized).

## Grouping

Grouping method is also known as pairs method. The main concept is making group to get the zero. The steps are:

If there is ax2 + bx + c then

1. Find number that added is equals with b and multiplied equals with c.
2. Write the addition to replace b. it will change three terms to be four terms.
3. Group / pair the polynomial. First-second terms and third-fourth terms.
4. Take the GCF on each pair. It is not always one term. It is possibly consisting of two terms.
5. Write down to be the factor of polynomial (multiple form).

If there is x2-x-12 then

x2-x-12 = x2-4x+3x-12                → -4+3 = -1 and (-4)(3) = 12

= (x)(x-4)+(3)(x-4)         → replace (-x)

= (x-4)(x+3)                    → take GCF and write as multiple form

## Four Terms Polynomials

This method is more complex than the previous. It is because the problem is polynomial that consists of four terms or the biggest degree is three. The steps are:

1. Split the polynomial into two parts
2. Determine the highest common factor in each part.
3. Write it out from the bracket
4. Regroup into multiple form.

If there is x3-3x2+2x-6 then

x3-3x2+2x-6 = (x3-3x2)+(2x-6) → split into two parts

= x2(x-3)+2(x-3) → determine highest common factor and write it out from bracket

= (x2+2)(x-3)       → regroup into multiple form

## Identities / Special Form

There are some identities that can be used to solve polynomial factorization problems.

1. (a+b)2 = a2 + 2ab + b2
2. (a-b)2 = a2 -2ab + b2
3. a2 – b2 = (a+b)(a-b)
4. a3+b3 = (a+b)(a2-ab+b2)
5. a3-b3 = (a-b)(a2+ab+b2)

Quadratic formula usually uses to determine solution of polynomial especially that has degree two (quadratic form). Another reason uses this method is because polynomial can be solve using others method or the solution is not integer.

If there is polynomial ax2+bx+c then the quadratic formula is

Using quadratic formula will get two solutions. if x1=m and x2=n, then the factor is (x-m) and (x-n).

## Examples

1. Factoring the polynomials:

a. 2x3 + 6x2
b. 12x3 + 2x2 – x
c. x2+4x-5
d. 2x2-2x+½

a. 2x3 + 6x2 = (2x2)(x)+(2x2)(3) = (2x2)(x+3)

b. 12x3 + 2x2 – x = (x)(12x2)+(2x2)(x)-(x)(1) = x (12x2+2x-1)

c. x2+4x-5 = x2+5x-x-5

= x2-x+5x-5

= (x)(x-1)+(5)(x-1)

= (x+5)(x-1)

d. 2x2-2x+½ = 2{x2-x+¼ }

=2{x2-½x-½x+¼}

=2{(x)(x-½)-(½)(x-½)

= 2(x- ½))(x-½)

2. Determine polynomial 2x2+kx-7 if one of the factor is (x-1) and determine the another.

Because of (x-1) is factor of 2x2+kx-7, based on factor theorem, then if substitute x = 1 will equals with zero.

x = 1 then 2(1)2+k(1)-7 = 0

2+k-7 = 0

k-5 = 0

k = 5

So, the polynomial is 2x2+5x-7.

Determining another factor can use division concept of polynomial or factoring polynomial.

Now, let’s try factoring polynomial.

2x2+5x-7 because of coefficient of x2 is not 1, then (2x+….)(x+….) as first step.

Then, determine factor of -7, they are:

• -1 & 7
• -7 & 1

Based on the both possibilities, determine which one that can be used to solve the problem.

It will be easy using trial & error.

•  -1 & 7 then (2x-1)(x+7) or (2x+7)(x-1)
• -7 & 1 then (2x-7)(x+1) or (2x+1)(x-7)

And the answer is (2x+7)(x-1). It is because one of the factors is given. Another way is trying one by one using multiplication concept of polynomial.

So, another factor is (2x+7).

3. Factor each of the following:

• a2 – ¼
• b2 – 5
• x2 + 7x + 12
• x2 – 3x – 18
• 8y2 – 16 – 28y
• x3 – x2 + 2x – 2
• 5x2y3 + 15x3y2
• 6t7 + 3t4 – 9t3
• a2 – ¼ = a2 – (½)2 (special form)

= (a- ½) (a + ½)

• b2 – 5 → b = 5 → b = ± √5

• x2 + 7x + 12 = x2 + 3x + 4x + 12

= (x2 + 3x) + (4x + 12)

= (x)(x+3) + (4)(x+3)

= (x+4)(x+3)

• x2 – 3x – 18 = x2 – 6x + 3x – 18

= (x2 – 6x) + (3x – 18)

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

= (x-6)(x+3)

• 8y2 – 16 – 28y = 4y (2y2 – 7y – 4)

= (4y)(2y+1)(y-4)

Note: 1 and -4 based on any two numbers that the multiplication result is c (factors of -4) and if the factors are checked by using multiplication then the result is 8y2 – 16 – 28y

• x3 – x2 + 2x – 2 = (x3 – x2) + (2x – 2)

= x2 (x – 1) + 2 (x-1)

= (x2+2) (x-1)

• 5x2y3 + 15x3y2 = (5x2y2) (y+3x)

6t7 + 3t4 – 9t3 = 3t3 (2t4 + t – 3)

= 3t3 {(2t+3)(t-1)}

= (3t2)(2t+3)(t-1)

4. Simple form of:

5. One of factor of polynomial is (2x+2). Then determine twice of another factor if 2x2 + mx + 2 is the polynomial.

• First step determine m.

2x+2 = 0 → x = -1

Then substitute to polynomial

2x2 + mx + 2 = 0

2(-1)2 + m(-1) + 2 = 0

2 – m + 2 = 0

m = 4

• Determine polynomial

2x2 + mx + 2 → 2x2 + 4x + 2

• Determine another factor

2x2 + 4x + 2 = (2x+2)(x+1)

*remember the step to determine the factors

• So, twice of another factor is 2(x+1) = 2x+2