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Q. 68

Precalculus Enhanced with Graphing Utilities
Found in: Page 210
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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Short Answer

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


The real solutions of the equation are -12, 2, and 4.

See the step by step solution

Step by Step Solution

Step 1. Finding the possible number of zeros 

The given equation is 2x3-11x2+10x+8=0

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

Step 2. Use the rational zero theorem 

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

The factors of the constant term 8 are

p: ±1, ±2, ±4, ±8

The factors of the leading coefficient 2 are

q: ±1, ±2

So, the possible rational zeros are

pq: ±1, ±2, ±4, ±8, ±12

Step 3. Finding the zero

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

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

Step 4. Finding the zero 

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 2x27x4.

Step 5. Use the zero product property 

As the remaining zeros satisfy the depressed equation



Use the property

2x+1=0x=-12 and x-4=0x=4

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