Question

which polynomial function has x-intercepts $-1$, $0$, and $2$ and passes through the point $(1, -6)$?
$\bigcirc$ $f(x) = x^3 - x^2 - 2x$
$\bigcirc$ $f(x) = 3x^3 - 3x^2 - 6x$
$\bigcirc$ $f(x) = x^3 + x^2 - 2x$
$\bigcirc$ $f(x) = 3x^3 + 3x^2 - 6x$

Explanation:

Step1: Form general polynomial

Given x-intercepts $x=-1, 0, 2$, the polynomial has factors $(x+1)$, $x$, and $(x-2)$. So the general form is:
$f(x) = ax(x+1)(x-2)$
where $a$ is a constant coefficient.

Step2: Expand the polynomial

First multiply the binomials:
$(x+1)(x-2) = x^2 - 2x + x - 2 = x^2 - x - 2$
Then multiply by $x$:
$x(x^2 - x - 2) = x^3 - x^2 - 2x$
So $f(x) = a(x^3 - x^2 - 2x)$

Step3: Solve for $a$

Use the point $(1, -6)$: substitute $x=1$, $f(x)=-6$
$-6 = a(1^3 - 1^2 - 2(1))$
$-6 = a(1 - 1 - 2)$
$-6 = a(-2)$
$a = \frac{-6}{-2} = 3$

Step4: Get final polynomial

Substitute $a=3$ into the expanded form:
$f(x) = 3(x^3 - x^2 - 2x) = 3x^3 - 3x^2 - 6x$

Answer:

B. $f(x) = 3x^3 - 3x^2 - 6x$