Question

algebra 2 362403 - 6 fy - 1
benchmark test 2
due feb 5 - 11:45 pm late
what are all the real and complex solutions of (x^{3}+2x^{2}+36x=-72)? round to the nearest tenth if necessary.
a. (2.8, 8.4i, -8.4i)
b. (-2, 6i, -6i)
c. (1.7, 1.7i, -5.1i)
d. (2, -6i)
question 20 of 30

Explanation:

Step1: Rewrite the equation

First, rewrite the given equation \(x^{3}+2x^{2}+36x=-72\) in standard form: \(x^{3}+2x^{2}+36x + 72 = 0\).

Step2: Factor by grouping

Group the terms: \((x^{3}+2x^{2})+(36x + 72)=0\).
Factor out the common factors from each group: \(x^{2}(x + 2)+36(x + 2)=0\).
Now, factor out \((x + 2)\): \((x + 2)(x^{2}+36)=0\).

Step3: Solve for \(x\)

Set each factor equal to zero:

• For \(x + 2 = 0\), we get \(x=-2\).
• For \(x^{2}+36 = 0\), we have \(x^{2}=-36\), so \(x=\pm\sqrt{-36}=\pm 6i\).

Answer:

B. \(-2, 6i, -6i\)