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Q. 68

Expert-verifiedFound in: Page 210

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

In Problems *63–72*, find the real solutions of each equation.

$2{x}^{3}-11{x}^{2}+10x+8=0$

The real solutions of the equation are $-\frac{1}{2},2,and4.$

The given equation is $2{x}^{3}-11{x}^{2}+10x+8=0$

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

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

The factors of the constant term $8$ are

$p:\pm 1,\pm 2,\pm 4,\pm 8$

The factors of the leading coefficient $2$ are

$q:\pm 1,\pm 2$

So, the possible rational zeros are

$\frac{p}{q}:\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$

Let's test the potential zero $1$ by using synthetic division

Since the remainder is not zero, it is not a factor.

Now, let's test the potential zero $2$ by using the synthetic division test

Since the remainder is zero, $(x-2)$ is a factor.

The depressed equation is $2{x}^{2}-7x-4.$

As the remaining zeros satisfy the depressed equation

So,

$2{x}^{2}-7x-4=0\phantom{\rule{0ex}{0ex}}2{x}^{2}-8x+x-4=0\phantom{\rule{0ex}{0ex}}2x(x-4)+1(x-4)=0\phantom{\rule{0ex}{0ex}}(2x+1)(x-4)=0$

Use the property

$2x+1=0\phantom{\rule{0ex}{0ex}}x=-\frac{1}{2}$ and $x-4=0\phantom{\rule{0ex}{0ex}}x=4$

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