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:
- (x-a) is a factr of p(x), if p(a) = 0
- If p(a) = 0 then (x-a) is a factor of p(x).
There are some factoring polynomial’s methods:
Table of Contents
Greatest Common Factor (GCF)
Remember about concept of 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.
- Find the GCF of all of terms in polynomial.
- Write / express each term as multiplying of GCF and the another.
- 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 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
- Find number that added is equals with b and multiplied equals with c.
- Write the addition to replace b. it will change three terms to be four terms.
- Group / pair the polynomial. First-second terms and third-fourth terms.
- Take the GCF on each pair. It is not always one term. It is possibly consisting of two terms.
- 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:
- Split the polynomial into two parts
- Determine the highest common factor in each part.
- Write it out from the bracket
- 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.
- (a+b)2 = a2 + 2ab + b2
- (a-b)2 = a2 -2ab + b2
- a2 – b2 = (a+b)(a-b)
- a3+b3 = (a+b)(a2-ab+b2)
- 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).
1. Factoring the polynomials:
a. 2x3 + 6x2
b. 12x3 + 2x2 – x
2. Determine polynomial 2x2+kx-7 if one of the factor is (x-1) and determine the another.
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
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.