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What is the product of all the possible values of x if

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by M7MBA » Wed May 16, 2018 1:34 am
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What is the product of all the possible values of x if $$x^2\left(x+2\right)+7x\left(x+2\right)+6\left(x+2\right)=0$$

A. –29
B. –12
C. 12
D. 29
E. 168

The OA is B.

Is there a simple way to solve this PS question? I would like someone give me a good approach here. Please. Thanks in advance.
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by GMATGuruNY » Wed May 16, 2018 2:01 am
M7MBA wrote:What is the product of all the possible values of x if $$x^2\left(x+2\right)+7x\left(x+2\right)+6\left(x+2\right)=0$$

A. –29
B. –12
C. 12
D. 29
E. 168
x²(x+2) + 7x(x+2) + 6(x+2) = 0.

Factor out x+2:
(x+2)(x² + 7x + 6) = 0

Factor the quadratic in red:
(x+2)(x+6)(x+1) = 0

Option for x:
x=-2, x=-6, x==-1.
Thus, the product of all the solutions = (-2)(-6)(-1) = -12.

The correct answer is B.
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by Jeff@TargetTestPrep » Thu May 17, 2018 5:19 pm
M7MBA wrote:What is the product of all the possible values of x if $$x^2\left(x+2\right)+7x\left(x+2\right)+6\left(x+2\right)=0$$

A. –29
B. –12
C. 12
D. 29
E. 168
We first factor the common (x + 2) from the left side. We have:

(x + 2)(x^2 + 7x + 6) = 0

Now, we can factor the quadratic x^2 + 7x + 6:

(x + 2)(x + 6)(x + 1) = 0

The three possible values for x are x = -2 or -6 or -1, for a product of -12.

Answer: B

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