Question

find all the zeros of the function (enter your answers as a comma - separated list.)
$h(x)=x^{4}+6x^{3}+10x^{2}+6x + 9$
$x = $
write the polynomial as a product of linear factors. use a graphing utility to verify your results graphically. (if possible, use the graphing utility to verify the imaginary zeros.)
$h(x) = $

Explanation:

Step1: Group terms for factoring

Group pairs of terms:
$(x^4 + 6x^3) + (10x^2 + 6x + 9)$

Step2: Factor out common terms

Factor $x^3$ from first group:
$x^3(x + 6) + (10x^2 + 6x + 9)$
Rewrite $10x^2$ as $x^2 + 9x^2$ to split:
$x^3(x + 6) + (x^2 + 6x + 9) + 9x^2$

Step3: Factor quadratic trinomial

Factor $x^2+6x+9$ as $(x+3)^2$:
$x^3(x + 6) + (x+3)^2 + 9x^2$
Rewrite $x^3(x+6)$ as $x^2(x(x+6)) = x^2(x^2+6x)$:
$x^2(x^2+6x) + 9x^2 + (x+3)^2 = x^2(x^2+6x+9) + (x+3)^2$
Factor $x^2+6x+9=(x+3)^2$:
$x^2(x+3)^2 + (x+3)^2 = (x+3)^2(x^2 + 1)$

Step4: Find zeros from factors

Set each factor equal to 0:

1. $(x+3)^2=0 \implies x=-3$ (double root)
2. $x^2+1=0 \implies x^2=-1 \implies x=\pm i$

Answer:

Zeros: $-3, -3, i, -i$
Product of linear factors: $h(x)=(x+3)^2(x-i)(x+i)$