 Factoring Higher Powers MathBitsNotebook.com Terms of Use   Contact Person: Donna Roberts Sometimes factoring problems with higher degree powers can be rewritten as a simpler factoring problem. 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 a2 with 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 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 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.                            (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 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. 