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^{ 2}y + 3xy^{ 2} + y^{ 3} = (x + y)^{ 3}
11: Perfect cube x^{ 3}  3x^{ 2}y + 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: February 2015
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