Sometimes factoring problems with higher degree powers can be rewritten as a simpler factoring problem.

expin1
 This first example replaces sections of the problem with a new variable.
This concept is referred to as factoring using substitution.
Factor: a4 + a2 - 12
At first glance, this problem doesn't look like anything we have seen. But upon closer examination, this is really a very simple problem that we already know how to solve. The trick is to look at the problem in a slightly different manner.

a4 + a2 - 12 = (a2)2 + (a2) - 12

If you now replace a2with x (or your favorite variable letter), the problem becomes very familiar.
Let a2 = x and substitute.        (a2)2 + (a2) - 12 = x2 + x - 12
Factor the new form:                 x2 + x - 12 = (x + 4)(x - 3)
Re-substitute x = a2.    (x + 4)(x - 3) = (a2 + 4)(a2 - 3) ANSWER

 

expin2
 This second example creates a new problem by finding a common factor.
Factor: 2x5 - 9x4 - 5x3
Look carefully! Each of these terms has a factor of x3. Factoring out x3 may create a problem which can be factored further.
                x3(2x2 - 9x - 5)
Factor the remaining trinomial using any method you wish.
Factor:     x3(2x2 - 9x - 5) = x3(2x + 1)(x - 5) ANSWER

 

expin3
 This third example has several solution methods.
Factor: m8 - 16
Since m8 and 16 are perfect squares, and this problem deals with subtraction, we are looking at factoring the difference of perfect squares.                 (m4 - 4)(m4 + 4)
But the first of the two factors is ANOTHER difference of perfect squares. Repeat the process.                 (m2 - 2)(m2 + 2)(m4 + 4) ANSWER
This problem can also be solved using the substitution approach, shown in example 1.
                           (m4)2 - 16
Let m4 = x.         (m4)2 - 16 = (x)2 - 16
Factor:                (x)2 - 16 = (x - 4)(x + 4)
Re-substitute:     (x - 4)(x + 4) = (m4 - 4)(m4 + 4)
Repeat process:  (m2)2 - 4
Let m2 = b.         (m2)2 - 4 = (b)2 - 4
Factor:                (b)2 - 4 = (b - 2)(b + 2)
Re-substitute:     (b - 2)(b + 2) = (m2 - 2)(m2 + 2) 
Total answer:      (m2 - 2)(m2 + 2)(m4 + 4) ANSWER

 

expin4
 This is a tough one!! What happens if there is a variable in the exponent?
Factor: x 2p + 2x p - 24
You have to be a bit more creative to see this solution. The secret is in the exponents.
x p • x p = x 2p making x 2p the square of x p. A substitution will solve our problem.
(x p )2 + 2(x p ) - 24
Let x p = a and substitute.        (x p)2 + 2(x p) - 24 = a2 + 2a - 24
Factor the new form:                 a2 + 2a - 24 = (a - 4)(a + 6)
Re-substitute a = x p.    (a - 4)(a + 6) = (x p - 4)(x p+ 6) ANSWER
NOTE: Without further information about "p", we do not know if further factoring is needed. If p were a positive even exponent, (x p - 4) could be the difference of two squares and could be factored further. However, without that information, we stop at our labeled answer.

 

divider

NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation
 and is not considered "fair use" for educators. Please read the "Terms of Use".

Topical Outline | Algebra 1 Outline | MathBitsNotebook.com | MathBits' Teacher Resources
Terms of Use    Contact Person: Donna Roberts

Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved.