Question

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

Answer 1

`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!