How do you solve #x^ { 3} + 3x ^ { 2} = 4#?

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Answer 1 · 2016-12-20T16:40:21

#{-2, 1}#

#x^3 + 3x^2 - 4 = 0#

We must check for rational factors using the rational root theorem and the remainder theorem.

By the rational root theorem, the possible zeros are:

#(+- 4, +- 2, +- 1)/(+-1)#

Let's now use remainder theorem to find which possible root, if any, will be a valid factor.

#1^3 + 3(1)^2 - 4 =^? 0#

#1 + 3 - 4 = 0" "color(green)(√)#

We have a factor!

By division (you can use long, or synthetic, whichever works for you), you should get the following:

enter image source here

This is equivalent to saying #x^3 - 3x^2 - 4 = (x - 1)(x^2 + 4x + 4) = 0#.

#x^2 + 4x + 4# is a perfect square, and can be factored as #(x + 2)(x + 2)# .

The final, factored equation is #(x - 1)(x +2)(x + 2) = 0#, with solutions #x= -2# and #x= 1#.

Hopefully this helps!