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Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248

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Chapter
Section
BuyFindarrow_forward

Applied Calculus

7th Edition
Waner + 1 other
ISBN: 9781337291248
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44.

x 3 6 x 2 + 12 x 8 = 0

To determine

To calculate: The all real solutions in the polynomial equation x36x2+12x8=0.

Explanation

Given Information:

The provided polynomial equation is x36x2+12x8=0.

Formula used:

Zero product rule:

An equation ab=0 is true if and only if a=0 or b=0, both

A product is zero if and only if at least one factor is zero

The value of x in the equation ax2+bx+c=0 can be given by the quadratic formula,

x=b±b24ac2a where a,b and c are real numbers with a0.

Calculation:

Consider the polynomial equation x36x2+12x8=0.

First determine the single solution in this cubic equation x36x2+12x8=0

Compare the equation x36x2+12x8=0 with the standard cubic equation ax3+bx2+cx+d=0.

a=1 and d=8

Therefore, possible factor of 1 is x=±1 and 8 is x=±1, x=±2, x=±4 and x=±8.

First try x=2 substitute in the provided polynomial x36x2+12x8=0, (2)36(2)2+12(2)8=?0              824+248=?00=0 which is true.

Therefore, x2 is a factor of the polynomial x36x2+12x8=0.

In order to find other two values of x, divide x36x2+12x8=0 by x2, x2x24x+4x36x2+12x8         x32x2_             4x2+12x              4x2+8x_

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