How do you find all the zeros of $F (x) = 5 x^{3} {(x + 3)}^{4} (x - 7)$ $F (x) = 5x^3 (x+ 3)^ 4 (x-7)$ with all its multiplicities?

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Answer 1 · 2017-08-20T10:06:27

The zeros are:

$x = 0$ $x=0" "$ with multiplicity $3$ $3$

$x = - 3$ $x=-3" "$ with multiplicity $4$ $4$

$x = 7$ $x=7" "$ with multiplicity $1$ $1$

Each linear factor corresponds to a zero. That is: $a$ $a$ is a zero if and only if $(x - a)$ $(x-a)$ is a factor.
The multiplicity of each factor is the multiplicity of the corresponding zero.

We can rewrite the given $F (x)$ $F(x)$ slightly to bring it into a form with factors with the form $(x - a)$ $(x-a)$ as follows:

$5 x^{3} {(x + 3)}^{4} (x - 7) = 5 {(x - 0)}^{3} {(x - (- 3))}^{4} (x - 7)$ $5x^3(x+3)^4(x-7) = 5(x-0)^3(x-(-3))^4(x-7)$

So the zeros are:

$x = 0$ $x=0" "$ with multiplicity $3$ $3$

$x = - 3$ $x=-3" "$ with multiplicity $4$ $4$

$x = 7$ $x=7" "$ with multiplicity $1$ $1$

How do you find all the zeros of F (x) = 5x^3 (x+ 3)^ 4 (x-7)F(x)=5x3(x+3)4(x−7)F (x) = 5x^3 (x+ 3)^ 4 (x-7) with all its multiplicities?