Rohan Mcpherson

2022-10-06

Find the P(x) polynomial of the smallest degree possible, with zeros 2 + i, and -3, such that -3 is a zero of multiplicity of 2 and P(-1)=2. Expand your polynomial completely.

### Answer & Explanation

Alannah Hanson

P(x) be apolynomial of the smallest degree 2+i is a zero of P(x)
$\left(2-i\right)$ is also a zero of p(x)
Also -3 is a zero of p(x) with multiplicity 2.
$p\left(x\right)=a\left(x+3{\right)}^{2}\left(x-2-i\right)\left(x-2+i\right)$
Here a is a constant
Given $p\left(-1\right)=2\phantom{\rule{0ex}{0ex}}⇒a\left(-1+3{\right)}^{2}\left(-1-2-i\right)\left(-1-2+i\right)=2\phantom{\rule{0ex}{0ex}}⇒4a\left(-3{\right)}^{2}-{i}^{2}=2\phantom{\rule{0ex}{0ex}}⇒4a\left(9+1\right)=2\phantom{\rule{0ex}{0ex}}⇒a=\frac{2}{40}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}\therefore P\left(x\right)=\frac{1}{20}\left(x+3{\right)}^{2}\left(x-2-i\right)\left(x-2+i\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{20}\left({x}^{2}+6x+9\right)\left({x}^{2}-4x+5\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{20}\left({x}^{4}+2{x}^{3}-10{x}^{2}-6x+15\right)$

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