FACTORING
The process of expressing a polynomial as a product is called factoring.
Many polynomial expressions can be written in simpler forms by factoring.
Basic types of factoring
Common monomial factor ab + ac = a(b+c)
Difference between two squares a2 - b2 = (a+b)(a-b)
Perfect square Trinomials a2 + 2ab + b2 = (a +b )2
a2 - 2ab + b2 = (a -b )2
Trinomials acx2 + (ad + bc)x + bd = (ax + b)(cx +
b)
Sum or Difference of Two Cubes a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a - b)(a2 + ab + b2)
Factoring by Grouping ac + bc + ad + bd = c(a+b) + d(a+b)
= (c+d)(a+b)
Factoring by Synthetic Division (present example)
Sum and Differences of Even or Odd xn - yn = (x-y)( xn-1+ xn-2y1 + xn-3y2 + … yn-1
Powers
Example 1: Factoring using Special Factoring Formulas
(a) 27x3 – 1 (b) x6 + 8
Solution:
(a) Using the Difference of Cubes Formula with a = 3x and b = 1, we get
27x3 – 1 = (3x)3 – 13
= (3x – 1)[(3x)2 + (3x)(1) + 12]
= (3x – 1)(9x2 + 3x + 1)
(b) Using the Sum of Cubes Formula with a = x2 and b = 2, we have
x6 + 8 = (x2)3 + 23
= (x2 + 2) (x4 – 2x2 + 4)
Examples 2: Factoring the Greatest Common Factor
Factor 6x3 y3 + 45x2 y2 + 21xy.
Solution:
The greatest common factor of the terms 6x3 y3, 45x2 y2 and 21xy is 3xy, so we
have
6x3 y3 + 45x2 y2 + 21xy = (3xy)(2x2 y2 + 15xy + 7)
Example 3 : Factoring a Trinomial with Leading Coefficient 1
Factor x2 + 2x – 15
Solution:
To factor a trinomial of the form x2 + bx + c, we note that
(x + r)(x + s) = x2 + (r + s)x + rs
so we need to choose numbers r and s so that r + s = b and rs = c.
Therefore, We need to find two integers whose product is -15 and whose sum is
2.
By trial and error we find that the two integers are -3 and 5. Thus, the
factorization is
x2 + 2x – 15 = (x − 3)(x + 5).
Example 4 : Factoring a Trinomial by Grouping
Factor 5x2 + 7x − 6 by grouping.
Solution:
To factor a trinomial in the form ax2 + bx + c by grouping, we find two numbers
with a product of ac and a sum of b. We use these numbers to divide the x term
into the sum of two terms and factor each portion of the expression separately,
then factor out the GCF of the entire expression.
We have a trinomial with a = 5, b = 7, and c = −6. First, determine ac = −30. We
need to find two numbers with a product of −30 and a sum of 7.
By trial and error we find that the two integers are -3 and 10. Thus,
5x2 − 3x + 10x − 6 Rewrite the original expression as ax2 + px + qx + c.
x(5x − 3) + 2(5x − 3) Factor out the GCF of each part.
(5x − 3)(x + 2) Ans. Factor out the GCF of the expression.
