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.

 

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