How do you solve using factoring: #4x^4 + 23x^2 = 72#?

1 Answer
Noah G
Oct 29, 2016

#x = +-3/2" AND "x = +- 2sqrt(2)i#

Explanation:

Rewrite in standard form, #ax^2 + bx + c = 0#.

#->4x^4 + 23x^2 - 72 = 0#

This is a fourth degree trinomial, so we can factor it using trinomial factoring rules, such as in #ax^2 + bx + c, a!=0, 1#, to factor you need to find two numbers that multiply to #ac# and that add to #b#.

#4x^4 + 32x^2 - 9x^2 - 72 = 0#

#4x^2(x^2 + 8) - 9(x^2 + 8) = 0#

#(4x^2 - 9)(x^2 + 8) = 0#

#x^2 = 9/4" AND "x^2 = -8#

#x = +-3/2" AND "x = +- 2sqrt(2)i#

Hopefully this helps!

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