How do you factor #s^ { 4} + 2s ^ { 2} - 63#?

2 Answers
Fleur
Jul 16, 2017

#(s^2+9)(s^2-7)#

Explanation:

#s^4+2s^2-63#

Find two numbers that have a product of #-63# and a sum of #2#. The numbers #9# and #-7# work. Also, since the first term is #s^4# and the second is #2s^2#, the first terms of the factored form should begin with #s^2#.

#(s^2+9)(s^2-7)#

Douglas K.
Jul 16, 2017

Let #x = s^2#, then:

#s^4+2s^2 - 63#

Becomes:

#x^2+2x-63#

This factors:

#(x + 9)(x -7)#

Reverse the substitution:

#(s^2+9)(s^2-7)#

This factors into:

#(s-3i)(s+3i)(s-sqrt7)(s+sqrt7)#

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