How do you find all the zeros of $F (x) = - 4 {(x + 7)}^{3} {(x - 7)}^{2}$ $F(x) = -4 (x+7)^3 (x-7)^2$ with all its multiplicities?

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Answer 1 · 2016-10-10T13:19:58

$F (x)$ $F(x)$ has zeros at $(- 7)$ $(-7)$ with multiplicity of $3$ $3$ and at $(+ 7$ $(+7$ with a multiplicity of $2$ $2$

$F (x) = - 4 {(x + 7)}^{3} (x - 7)$ $F(x)=-4(x+7)^3(x-7)$ ^2#

A term of $(x + 7)$ $(x+7)$ implies a zero at $x = - 7$ $x=-7$
A term of $(x + 7)$ $(x+7)$ implies a zero at $(x = + 7$ $(x=+7$

$F(x)=-4 * underbrace((x+7)(x+7)(x+7))_"multiplicity of 3" * underbrace((x-7)(x-7))_"multiplicity of 2"$

How do you find all the zeros of F(x) = -4 (x+7)^3 (x-7)^2F(x)=−4(x+7)3(x−7)2F(x) = -4 (x+7)^3 (x-7)^2 with all its multiplicities?