How do you find the zeros with multiplicity for the function $p (x) = (x^{3} - 8) (x^{5} - 4 x^{3})$ $p(x)=(x^3-8)(x^5-4x^3)$ ?

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How do you find the zeros with multiplicity for the function $p (x) = (x^{3} - 8) (x^{5} - 4 x^{3})$ $p(x)=(x^3-8)(x^5-4x^3)$ ?

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Answer 1 · 2016-11-15T23:34:55

$p (x)$ $p(x)$ has zeros:

$0$ $0$ with multiplicity $3$ $3$

$2$ $2$ with multiplicity $2$ $2$

$- 2$ $-2$ with multiplicity $1$ $1$

$- 1 + \sqrt{3} i$ $-1+sqrt(3)i$ with multiplicity $1$ $1$

$- 1 - \sqrt{3} i$ $-1-sqrt(3)i$ with multiplicity $1$ $1$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

The difference of cubes identity can be written:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

We find:

$p (x) = (x^{3} - 8) (x^{5} - 4 x^{3})$ $p(x) = (x^3-8)(x^5-4x^3)$

$p (x) = (x^{3} - 2^{3}) x^{3} (x^{2} - 2^{2})$ $color(white)(p(x)) = (x^3-2^3)x^3(x^2-2^2)$

$p (x) = (x - 2) (x^{2} + 2 x + 4) x^{3} (x - 2) (x + 2)$ $color(white)(p(x)) = (x-2)(x^2+2x+4)x^3(x-2)(x+2)$

$p (x) = x^{3} {(x - 2)}^{2} (x + 2) (x^{2} + 2 x + 4)$ $color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+4)$

$p (x) = x^{3} {(x - 2)}^{2} (x + 2) (x^{2} + 2 x + 1 + 3)$ $color(white)(p(x)) = x^3(x-2)^2(x+2)(x^2+2x+1+3)$

$p (x) = x^{3} {(x - 2)}^{2} (x + 2) ({(x + 1)}^{2} - {(\sqrt{3} i)}^{2})$ $color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)^2-(sqrt(3)i)^2)$

$p (x) = x^{3} {(x - 2)}^{2} (x + 2) ((x + 1) - \sqrt{3} i) ((x + 1) + \sqrt{3} i)$ $color(white)(p(x)) = x^3(x-2)^2(x+2)((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)$

$p (x) = x^{3} {(x - 2)}^{2} (x + 2) (x + 1 - \sqrt{3} i) (x + 1 + \sqrt{3} i)$ $color(white)(p(x)) = x^3(x-2)^2(x+2)(x+1-sqrt(3)i)(x+1+sqrt(3)i)$

Hence zeros:

$0$ $0$ with multiplicity $3$ $3$

$2$ $2$ with multiplicity $2$ $2$

$- 2$ $-2$ with multiplicity $1$ $1$

$- 1 + \sqrt{3} i$ $-1+sqrt(3)i$ with multiplicity $1$ $1$

$- 1 - \sqrt{3} i$ $-1-sqrt(3)i$ with multiplicity $1$ $1$

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How do you find the zeros with multiplicity for the function $p (x) = (x^{3} - 8) (x^{5} - 4 x^{3})$ $p(x)=(x^3-8)(x^5-4x^3)$ ?

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Explanation:

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How do you find the zeros with multiplicity for the function p(x)=(x3−8)(x5−4x3)p(x)=(x^3-8)(x^5-4x^3)?

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How do you find the zeros with multiplicity for the function $p (x) = (x^{3} - 8) (x^{5} - 4 x^{3})$ $p(x)=(x^3-8)(x^5-4x^3)$ ?