Question

Using the Factor Theorem, which of the polynomial functions has the zeros 4, $\sqrt7$, and $-\sqrt{7\\}$?
f (x) = x3 – 4x2 + 7x + 28
f (x) = x3 – 4x2 – 7x + 28
f (x) = x3 + 4x2 – 7x + 28
f (x) = x3 + 4x2 – 7x – 28

Answer 1

Answer:

B

Step-by-step explanation:

According to the Factor Theorem, if (x - a) is a factor (where a is a zero) of the polynomial P(x), then P(a) must equal zero.

Our zeros are 4, √7, and - √7. Hence, when evaluating P(4), P(√7), and P(-√7), all must evaluate to zero.

Testing each choice, we can see that only choice B is true. That is:

$\displaystyle \displaystyle \begin{aligned} f(4)&= (4)^3 - 4(4)^2 - 7(4) + 28 \\ &= (64) - (64) - (28) + 28 \\ &= 0 \stackrel{\checkmark}{=}0 \\f(\sqrt{7}) &= (\sqrt{7})^3 - 4(\sqrt{7})^2 - 7(\sqrt{7}) + 28 \\ &= (7\sqrt{7}) - (28) - (7\sqrt{7}) + 28 \\ &= 0 \stackrel{\checkmark}{=} 0\\ f(-\sqrt{7}) &= (-\sqrt{7})^3 - 4(-\sqrt{7})^2 - 7(-\sqrt{7}) + 28 \\ &= (-7\sqrt{7}) - (28) + (7\sqrt{7}) + 28 \\ &= 0 \stackrel{\checkmark}{=} 0 \end{aligned}$

In conclusion, our answer is B.