How do you find all the zeros of $F (x) = 8 (x - 6) {(x + 6)}^{5}$ $F(x) = 8(x-6)(x+6)^5$ with all its multiplicities?

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Answer 1 · 2018-01-23T17:18:32

We have two zeros $6$ $6$ with multilicity of $1$ $1$ and $- 6$ $-6$ with multiplicity of $5$ $5$ .

Zeros of $f (x) = k {(x - α)}^{a} {(x - β)}^{b} {(x - γ)}^{c}$ $f(x)=k(x-alpha)^a(x-beta)^b(x-gamma)^c$

are $α$ $alpha$ with multiplicity of $a$ $a$ , $β$ $beta$ with multiplicity of $b$ $b$ and $γ$ $gamma$ with multipicity of $c$ $c$ .

Hence in $f (x) = 8 (x - 6) {(x + 6)}^{5}$ $f(x)=8(x-6)(x+6)^5$

we have two zeros $6$ $6$ with multilicity of $1$ $1$ and $- 6$ $-6$ with multiplicity of $5$ $5$ .

How do you find all the zeros of F(x)=8(x−6)(x+6)5F(x) = 8(x-6)(x+6)^5 with all its multiplicities?