How do you write a polynomial in standard form given zeros -1 (multiplicity 2), -2 - i (multiplicity 1)?

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How do you write a polynomial in standard form given zeros -1 (multiplicity 2), -2 - i (multiplicity 1)?

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Answer 1 · 2016-09-29T02:46:24

$f (x) = x^{4} + 6 x^{3} + 14 x^{2} + 14 x + 5$ $f(x)=x^4+6x^3+14x^2+14x+5$

Explanation:

Zeros:
$- 1$ $-1$ multiplicity 2
$(- 2 - i)$ $(-2-i)$ multiplicity 1

If the zero is $- 1$ $-1$ , the factor is $(x - (- 1)) = (x + 1)$ $(x-(-1))=(x+1)$ .
A multiplicity of $2$ $color(red)2$ implies there are $2$ $color(red)2$ factors $(x + 1)$ $(x+1)$ or $(x + 1) (x + 1) = {(x + 1)}^{2} = (x^{2} + 2 x + 1)$ $(x+1)(x+1)= (x+1)^color(red)2 =(x^2+2x+1)$

For the complex zero $(- 2 - i)$ $(-2-i)$ , there will also be a zero at the complex conjugate $(- 2 + i)$ $(-2+i)$

The factors are $(x - (- 2 - i))$ $(x-(-2-i))$ and $(x - (- 2 + i))$ $(x-(-2+i))$
or
$(x + 2 + i) (x + 2 - i)$ $(x+2+i)(x+2-i)$
$(x^{2} + 2 x - i x + 2 x + 4 - 2 i + i x + 2 i - i^{2})$ $(x^2+2x-ix+2x+4-2i+ix+2i-i^2)$
$(x^{2} + 4 x + 4 - (- 1))$ $(x^2+4x+4-(-1))$
$(x^{2} + 4 x + 5)$ $(x^2+4x+5)$

Multiply all factors to find the polynomial in standard form
$f (x) = (x^{2} + 2 x + 1) (x^{2} + 4 x + 5) =$ $f(x)=(x^2+2x+1)(x^2+4x+5)=$

$x^{4} + 4 x^{3} + 5 x^{2}$ $x^4+4x^3+5x^2$
$a a a a 2 x^{3} + 8 x^{2} + 10 x$ $color(white)(aaaa)2x^3+8x^2+10x$
$a a a a a a a a a a x^{2} + 4 x + 5 =$ $color(white)(aaaaaaaaaa)x^2+4x+5=$

$f (x) = x^{4} + 6 x^{3} + 14 x^{2} + 14 x + 5$ $f(x)=x^4+6x^3+14x^2+14x+5$

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How do you write a polynomial in standard form given zeros -1 (multiplicity 2), -2 - i (multiplicity 1)?

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Explanation:

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