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Two zeros of f(x)=x^3-6x^2-16x+96 are 4 and -4. is the third zero real or imaginary?

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Answer: Real.

Explanation:

The third zero, or solutions/roots, is a real zero. It is 6.

To find the answer to this, we will solve by graphing. See attached. The graph intercepts the x-axis, when y = 0 ("zeros") three times. These points have x-values of 4, -4, and our last one of 6.

We can substitute this back into the equation and see that it is true.

f(x) = x³ - 6x² - 16x + 96

f(6) = (6)³ - 6(6)² - 16(6) + 96

0 = 216 - 216 - 96 + 96

0 = 0

Two zeros of f(x)=x^3-6x^2-16x+96 are 4 and -4. is the third zero real or imaginary-example-1
User Jay Supeda
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