Tutorial with detailed solutions on factoring polynomials.

## Factoring Polynomials Formulas1: One common factor. a x + a y = a (x + y) 2: Sevearl grouped common factor. a x + a y + b x + b y = a(x + y) + b(x + y) = (a + b ) (x + y) 3: Difference of two squares (1). x ^{ 2} - y^{ 2} = (x + y)(x - y)
4: Difference of two squares (2). (x + y) ^{ 2} - z^{ 2} = (x + y + z)(x + y - z)
5: Sum of two cubes. x ^{ 3} + y^{ 3} = (x + y)(x^{ 2} - x y + y^{ 2})
6: Difference of two cubes. x ^{ 3} - y^{ 3} = (x - y)(x^{ 2} + x y + y^{ 2})
7: Difference of fourth powers. x ^{ 4} - y^{ 4} = (x^{ 2} - y^{ 2})(x^{ 2} + y^{ 2}) = (x + y)(x - y)(x^{ 2} + y^{ 2})
8: Perfet square x ^{ 2} + 2xy + y^{ 2} = (x + y)^{ 2}
9: Perfet square x ^{ 2} - 2xy + y^{ 2} = (x - y)^{ 2}
10: Perfect cube x ^{ 3} + 3x^{ 2}y + 3xy^{ 2} + y^{ 3} = (x + y)^{ 3}
11: Perfect cube x ^{ 3} - 3x^{ 2}y + 3xy^{ 2} - y^{ 3} = (x - y)^{ 3}
## Examples in Factoring Polynomials with Solutions## Example 1Factor the binomial 9 - 4x^{ 2}
Solution
Rewrite the given expression as the difference of two squares then apply formula 1 given above. 9 - 4x ^{ 2} = 3^{ 2} - (2x)^{ 2} = (3 - 2x)(3 + 2x)
As a practice, multiply (3 - 2x)(3 + 2x) to obtain the given expression. ## Example 2Factor the trinomial 9x^{ 2} + 3x - 2
Solution
To factor the above trinomial, we need to write it in the form. 9x ^{ 2} + 3x - 2 = (ax + m)(bx + n)
Expand the product on the right above 9x ^{ 2} + 3x - 2 = abx^{ 2} + x(mb + na) + mn
For the polynomial on the left to be equal to the polynomial on the right we need to have equal corresponding coefficients, hence ab = 9 mb + na = 3 mn = -2 Trial values for a and b are: a = 1 and b = 9 or a = 3 and b = 3 Trial values for m and n are: m = 1 and n = -2, m = 2 and n = -1, m = -1 and n = 2 and m = -2 and n = 1. Trying various values for a, b, m and n among the list above, we arrive at: 9x ^{ 2} + 3x - 2 = (3x + 2)(3x - 1)
As a practice, multiply (3x + 2)(3x - 1) and simplify to obtain the given trinomial. ## Example 3Factor the polynomial x^{ 3} + 2x^{ 2} - 16x - 32
Solution
Group terms that have common factors. x ^{ 3} + 2x^{ 2} - 16x - 32 = (x^{ 3} + 2x^{ 2}) - (16x + 32)
Factor the grouped terms = x ^{ 2}(x + 2) - 16(x + 2)
Factor x + 2 out = (x + 2)(x ^{ 2} - 16)
The term (x ^{ 2} - 16) is the difference of two squares and can be factored using formula 1 above
= (x + 2)(x + 4)(x - 4) Check the above answer by expanding the obtained result.
## ExercisesFactor the polynomials.1: (x + 1) ^{ 2} - 4
2: x ^{ 2} + 5x + 6
3: x ^{ 3} - 1
4: x ^{ 3} - x^{ 2} - 25x + 25
Solutions to above exercises
1: (x + 1) ^{ 2} - 4 = (x - 1)(x + 3)
2: x ^{ 2} + 5x + 6 = (x + 2)(x + 3)
3: x ^{ 3} - 1 = (x - 1)(x^{ 2} + x + 1)
4: x ^{ 3} - x^{ 2} - 25x + 25 = (x - 1)(x + 5)(x - 5)
## More references and links to polynomial functionsMultiplicity of Zeros and Graphs Polynomials.Find Zeros of Polynomial Functions - Problems Polynomial Functions, Zeros, Factors and Intercepts |