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Q. 55

Expert-verifiedFound in: Page 209

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

In Problems *39–56*, find the real zeros of *f*. Use the real zeros to factor *f*.

$f\left(x\right)=4{x}^{5}-8{x}^{4}-x+2$

The real zeros of *f *are $2,-\frac{1}{\sqrt{2}},and\frac{1}{\sqrt{2}}.$

The factored form of *f *is $(2{x}^{2}+1)(\sqrt{2}x-1)(\sqrt{2}x+1)(x-2).$

The given function is $f\left(x\right)=4{x}^{5}-8{x}^{4}-x+2$

The given function is of degree five, so it has at most five real zeros.

As all the coefficients are integers so we use the rational zeros theorem.

The factors of the constant term $2$ are

$p:\pm 1,\pm 2$

The factors of the leading coefficient $4$ are

$q:\pm 1,\pm 2,\pm 4$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm \frac{1}{2},\pm \frac{1}{4}$

The graph is

From the graph, we conclude that it has three roots.

Since $2$ appears to be zero and a potential rational zero also.

By evaluating we get, $f\left(2\right)=0$

Thus, $(x-2)$ is a factor of *f*.

Use synthetic division to factor *f*

So, factor *f *is localid="1646140146826" $(x-2)(4{x}^{4}-1)$

Put depressed equation to zero and factor by grouping

$4{x}^{4}-1=0\phantom{\rule{0ex}{0ex}}4{x}^{4}=1\phantom{\rule{0ex}{0ex}}{x}^{4}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}x=\pm \frac{1}{\sqrt{2}}$

Therefore, the other zeros are $-\frac{1}{\sqrt{2}}and\frac{1}{\sqrt{2}}.$

The factored form of *f * is $(2{x}^{2}+1)(\sqrt{2}x-1)(\sqrt{2}x+1)(x-2).$

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