Factor Polynomials

Tutorial with detailled solutions on factoring polynomials.

Factoring Formulas

1:    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 2y + 3xy 2 + y 3 = (x + y) 3

11:     Perfect cube   x 3 - 3x 2y + 3xy 2 - y 3 = (x - y) 3


Example 1: Factor 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 2: Factor 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 3: Factor 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.
Exercises: Factor 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 functions.
Multiplicity of Zeros and Graphs Polynomials.

Find Zeros of Polynomial Functions - Problems

Polynomial Functions, Zeros, Factors and Intercepts


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Updated: 2 April 2013

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