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