Darian Hubbard

2022-07-28

Find all real zeros of the polynomial. use quadratic formula. $P\left(x\right)={x}^{5}-8{x}^{4}+7{x}^{3}+30{x}^{2}+2x-12$

Franklin Frey

Expert

Consider the polynomial $P\left(x\right)={x}^{5}-8{x}^{4}+7{x}^{3}+30{x}^{2}+2x-12$
Factor the polynomial as
$P\left(x\right)={x}^{3}\left({x}^{2}-8{x}^{2}+7\right)+30{x}^{2}+2x-12\phantom{\rule{0ex}{0ex}}=\left(x+1{\right)}^{2}\left(x-6\right)\left({x}^{2}-4x+2\right)$
Now to find real zeros, set P(x)=0 and solve for x.
$P\left(x\right)=0\phantom{\rule{0ex}{0ex}}\left(x+1{\right)}^{2}\left(x-6\right)\left({x}^{2}-4x+2\right)=0\phantom{\rule{0ex}{0ex}}\left(x+1{\right)}^{2}=0\phantom{\rule{0ex}{0ex}}x=-1$
Now solve the equation ${x}^{2}-4x+2=0$ using quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ as
$x=\frac{\left(-4\right)±\sqrt{\left(-4{\right)}^{2}-4\left(1\right)\left(2\right)}}{2\left(1\right)}\phantom{\rule{0ex}{0ex}}=\frac{4±\sqrt{8}}{2}\phantom{\rule{0ex}{0ex}}=\frac{4±2\sqrt{2}}{2}\phantom{\rule{0ex}{0ex}}=2±\sqrt{2}$
Therefore, the real zeros are $\left[-1,6,2-\sqrt{2},2+\sqrt{2}\right]$

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