# How to Factoring Polynomials Worksheet | Algebra

In Algebra, Polynomial Factoring is the process of expressing a polynomial equation as a product of two or more similar polynomials. Each polynomial as a product of the given polynomial equation. For example (x - 2) and (x + 2) are the factors of x2 - 4
x2 - 4 = (x - 2) (x + 2)
Thus factoring is mainly used for solving or simplifying polynomial expressions. The process of polynomial factoring is also called as the resolution into factors.

How to Find Factoring of Polynomial:
There are three methods (common factors, grouping terms and using formulas) are used to find polynomial factors in common

Factoring Polynomials by Common Factor:
First we have to find the common factor for the given polynomial equation. If a given algebraic polynomial equation A contains common factor B, then divide each term of equation A by common factor B and you will get an expression C. Now B and C are the factors of A and it can be expressed as A = B x C

Example: Find the factors for the polynomial equation 7x5y3 + 4 x2y3 + 10 xy3?
Solution:
In the above equation xy3 is common factor.
7x5y3 + 4 x2y3 + 10 xy3 = xy3 (7x5y3/xy3 + 4 x2y3/xy3 + 10 xy3/xy3)
7x5y3 + 4 x2y3 + 10 xy3 = xy3 (7x4+4x+10)

Try Yourself
Find the polynomial Factors for the below equations
1. 9x - 3y
2. 4x3 - 8x2 + 16x
3. p5 + 4p
4. 6a5b5 + 3a2b3 + 14ab3
5. 7ab - 21a2b22

Factoring Polynomials by Grouping terms
If the Polynomial Equation don't have common factors then group the terms. By grouping terms in a appropriate manner you can get a common factor.
Example: Find polynomial factors for the below equation
x2 +x - 2xy - 2y
Solution:
x2 +x - 2xy - 2y = x2 - 2xy +x - 2y
Take X as common for first two terms
= x(x - 2y) + (x - 2y)
= (x - 2y) (x + 1)
So the factors for x2 +x - 2xy - 2y are (x - 2y) and (x + 1)

Try Yourself
Find the polynomial Factors for the below equations
1. ab - 2c - cb + 2a
2. a3 - 2a2 - 2a + 4
3. a3 - a2 - ba + b
4. 2x3 - x2 + 2x - 1
5. 8a3 + 4a2 + 4a + 2

Polynomial Factors by Formulas
In some cases, the below factorization formulas can be used to resolving a polynomial into factors.
1. (x + y)2 = x2 + y2 + 2xy
2. (x - y)2 = x2 + y2 - 2xy
3. (x + y) (x - y) = x2 - y2
4. (x + y) (x2 - xy + y2) = x3 - y3
5. (x - y) ( x2 + xy + y2) = x3 - y3
6. (x + y)2 = x3 + y3 + 3x2y + 3y2x = x3 + y3 + 3xy (x + y)
7. (x - y)3 = x3 - y3 - 3x2y + 3y2x = x3 - y3 - 3xy (x - y)
8. (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
9. (x + y + z)(x2 + y2 + z2 - xy - yz - zx) = x3 + y3 + z3 - 3xyz

Examples Problems
1. Find factors of the equation 4a2 + 12ab + 9b2
Solution:
The given polynomial equation can be written as follows
4a2 + 12ab + 9b2 = (2a)2 + 2 (2a)(3b) + (3b)2
The above equation can be simplified by using (x + y)2
= (2a + 3b)2

2. Find factors of the equation 125a3 + 64b3
Solution:
125a3 = (5a)3
64b3 = (4b)3
lets consider x = 5a and y = 4b
125a3 + 64b3 = (5a)3 + (4b)3
x3 + y3 = (x + y)(x2 +y2 - xy)
= (5a + 4b)[(5a)2 + (4b)2 - (5a)(4b)]
= (5a + 4b)[25a2 + 16b2 - 20ab]

3. Factorize 6(a-1)2b - 5(a - 1)b2 - 6b3
Solution:
put x= a-1 in the above equation,
6(a-1)2b - 5(a - 1)b2 - 6b3 = 6(x)2b - 5(x)b2 - 6b3
= b(6x2 - 5bx - 6b2)
Here we find 6 X -6 = -36 = -9 X 4 and -9 + 4 = -5
= b(6x2 - 9bx + 4bx - 6b2)
= b[3x(2x - 3b) + 2b(2x - 3b)]
= b(2x - 3b)(3x + 2b)
= b[2(a-1) - 3b] [3(a - 1) + 2b]
= b[(2a - 3b -2)(3a + 2b - 3)]
There are quadratic with integer coefficients which can not be factored with integer coefficients

Try Yourself:
Find the polynomial factors for the below equations
1. x4 + x2y2 + y4
2. 4a2 + 20ab + 25b2 - 10a - 25b
3. 27a3 + b3 + 27a2b + 9ab2
4. a3 - b3 + 1 + 3ab
5. 3√3a3b3 + 27c3