Answer
$3$ is a zero of multiplicity $3$.
$\displaystyle \frac{1}{2}$ is a zero of multiplicity $3.$
$2+\sqrt{5}$ is a zero of multiplicity $1.$
Work Step by Step
$x=k$ is a zero of a polynomial function $f(x)$, then $f(k)=0..$
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
The number of times $(x-k)$ occurs as a factor is referred to as the multiplicity of the zero.
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$(2x^{2}-7x+3)^{3}=$ ... factor the trinomial,
... factors of $ac=6$ whose sum is $b=-7$ ... are $-6$ and $-1$.
... rewrite $bx$ and factor in pairs
$=[2x^{2}-6x-x+3]^{3}$
$=[2x(x-3)-(x-3)]^{3}$
$=[(x-3)(2x-1)]^{3}$
$=[(x-3)\displaystyle \cdot 2(x-\frac{1}{2})]^{3}$
$=(x-3)^{3}\displaystyle \cdot 2^{3}(x-\frac{1}{2})^{3}$
$f(x)=8(x-3)^{3}(x-\displaystyle \frac{1}{2})^{3}\cdot[x-(2+\sqrt{5})]$
$(x-3)^{3}$ is a factor of f $\Rightarrow k_{1}=3$ is a zero of multiplicity $3$.
$(x-\displaystyle \frac{1}{2})^{3}$ is a factor of f $\Rightarrow k_{2}=1$ is a zero of multiplicity $3.$
$[x-(2+\sqrt{5})]$ is a factor of f $\Rightarrow k_{3}=2+\sqrt{5}$ is a zero of multiplicity $1.$
