3 years ago

URGENT, HERES THE EQUATION!:)

Mathematics

1 answer:

aev [14]3 years ago

4 0

The answer to the question would be 3.5

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There are 3 zeros, which are:

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Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14$

To get the zeros of $f(x)$ , set $y$ or $f(x) =0$

$0=x^3-7x^2-2x+14$

$x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$

$\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

Using the zero factor principle:

$\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0$

solving

$x-7=0$

$x=7$

$x+\sqrt{2}=0$

$x=-\sqrt{2}$

$x-\sqrt{2}=0$

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Therefore, there are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

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