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

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

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Answer 1 · 2016-10-29T14:49:20

$x = \pm \frac{3}{2} AND x = \pm 2 \sqrt{2} i$ $x = +-3/2" AND "x = +- 2sqrt(2)i$

Explanation:

Rewrite in standard form, $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ .

$\to 4 x^{4} + 23 x^{2} - 72 = 0$ $->4x^4 + 23x^2 - 72 = 0$

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

$4 x^{4} + 32 x^{2} - 9 x^{2} - 72 = 0$ $4x^4 + 32x^2 - 9x^2 - 72 = 0$

$4 x^{2} (x^{2} + 8) - 9 (x^{2} + 8) = 0$ $4x^2(x^2 + 8) - 9(x^2 + 8) = 0$

$(4 x^{2} - 9) (x^{2} + 8) = 0$ $(4x^2 - 9)(x^2 + 8) = 0$

$x^{2} = \frac{9}{4} AND x^{2} = - 8$ $x^2 = 9/4" AND "x^2 = -8$

$x = \pm \frac{3}{2} AND x = \pm 2 \sqrt{2} i$ $x = +-3/2" AND "x = +- 2sqrt(2)i$

Hopefully this helps!

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

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Explanation:

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