Question

Which of the following is a factor of the polynomial x³+x²-17x+15?

• A. x+3
• B. x-3
• C. 2x+3
• D. 2x-3

Answer 1

The Correct Option is B

Step 1: Recall the Factor Theorem.

The Factor Theorem states that if $(x - c)$ is a factor of a polynomial $P(x)$, then $P(c) = 0$. To check if any of the given options is a factor, we substitute the corresponding value of $c$ into $P(x)$ and verify if $P(c) = 0$.

Step 2: Test each option.

(1) $x + 3$:

If $x + 3$ is a factor, then $c = -3$. Substitute $x = -3$ into $P(x)$:

$$P(-3) = (-3)^3 + (-3)^2 - 17(-3) + 15 = -27 + 9 + 51 + 15 = 48.$$

Since $P(-3) \neq 0$, $x + 3$ is not a factor.

(2) $x - 3$:

If $x - 3$ is a factor, then $c = 3$. Substitute $x = 3$ into $P(x)$:

$$P(3) = (3)^3 + (3)^2 - 17(3) + 15 = 27 + 9 - 51 + 15 = 0.$$

Since $P(3) = 0$, $x - 3$ is a factor.

(3) $2x + 3$:

If $2x + 3$ is a factor, then $c = -\frac{3}{2}$. Substitute $x = -\frac{3}{2}$ into $P(x)$:

$$P\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^3 + \left(-\frac{3}{2}\right)^2 - 17\left(-\frac{3}{2}\right) + 15.$$

Simplify step-by-step:

$$P\left(-\frac{3}{2}\right) = -\frac{27}{8} + \frac{9}{4} + \frac{51}{2} + 15.$$

Convert all terms to have a denominator of 8:

$$P\left(-\frac{3}{2}\right) = -\frac{27}{8} + \frac{18}{8} + \frac{204}{8} + \frac{120}{8}.$$

$$P\left(-\frac{3}{2}\right) = \frac{-27 + 18 + 204 + 120}{8} = \frac{315}{8}.$$

Since $P\left(-\frac{3}{2}\right) \neq 0$, $2x + 3$ is not a factor.

(4) $2x - 3$:

If $2x - 3$ is a factor, then $c = \frac{3}{2}$. Substitute $x = \frac{3}{2}$ into $P(x)$:

$$P\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^3 + \left(\frac{3}{2}\right)^2 - 17\left(\frac{3}{2}\right) + 15.$$

Simplify step-by-step:

$$P\left(\frac{3}{2}\right) = \frac{27}{8} + \frac{9}{4} - \frac{51}{2} + 15.$$

Convert all terms to have a denominator of 8:

$$P\left(\frac{3}{2}\right) = \frac{27}{8} + \frac{18}{8} - \frac{204}{8} + \frac{120}{8}.$$

$$P\left(\frac{3}{2}\right) = \frac{27 + 18 - 204 + 120}{8} = \frac{-39}{8}.$$

Since $P\left(\frac{3}{2}\right) \neq 0$, $2x - 3$ is not a factor.

Final Answer: The correct factor is $\mathbf{x - 3}$, which corresponds to option $\mathbf{(2)}$.