Question

select the correct answer.
which function has exactly three distinct real zeros?
a. $h(x)=(x-9)^2(x-4)^2$
b. $h(x)=x(x+7)^2$
c. $h(x)=(x-3)(x+1)(x+3)(x+8)$
d. $h(x)=(x-2)^2(x+4)(x-1)$

Explanation:

Step1: Identify zeros for Option A

Set $h(x)=0$: $(x-9)^2(x-4)^2=0$. Zeros are $x=9$ (multiplicity 2), $x=4$ (multiplicity 2). Distinct zeros: 2.

Step2: Identify zeros for Option B

Set $h(x)=0$: $x(x+7)^2=0$. Zeros are $x=0$, $x=-7$ (multiplicity 2). Distinct zeros: 2.

Step3: Identify zeros for Option C

Set $h(x)=0$: $(x-3)(x+1)(x+3)(x+8)=0$. Zeros are $x=3, x=-1, x=-3, x=-8$. Distinct zeros: 4.

Step4: Identify zeros for Option D

Set $h(x)=0$: $(x-2)^2(x+4)(x-1)=0$. Zeros are $x=2$ (multiplicity 2), $x=-4, x=1$. Distinct zeros: 3.

Answer:

D. $h(x) = (x - 2)^2(x + 4)(x - 1)$