Question

19 part a
(a) based on the graph, how many zeroes does the polynomial function, $f(x) = -x^4 - 3x^3 + 6x^2 + 8x$ have?
answer:
part b
(b) what are the zeroes of the above polynomial function?

Explanation:

Step1: Count x-axis intersections

The graph crosses the x-axis at 1 point.

Step2: Factor the polynomial

Factor $f(x) = -x^4 -3x^3 +6x^2 +8x$:
First factor out $-x$:
$f(x) = -x(x^3 +3x^2 -6x -8)$
Use rational root theorem to find a root of $x^3 +3x^2 -6x -8$: $x=2$ is a root. Factor out $(x-2)$ via polynomial division:
$x^3 +3x^2 -6x -8 = (x-2)(x^2 +5x +4)$
Factor the quadratic: $x^2 +5x +4=(x+1)(x+4)$
So $f(x) = -x(x-2)(x+1)(x+4)$

Step3: Solve for zeros

Set $f(x)=0$:
$-x(x-2)(x+1)(x+4)=0$
Solve each factor:
$-x=0 \implies x=0$
$x-2=0 \implies x=2$
$x+1=0 \implies x=-1$
$x+4=0 \implies x=-4$

Answer:

(a) 1
(b) $x=-4$, $x=-1$, $x=0$, $x=2$