   Chapter 0.5, Problem 34E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. y 3 + 64 = 0

To determine

To calculate: The all real solutions in the polynomial equation y3+64=0.

Explanation

Given Information:

The provided polynomial equation is y3+64=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 y3+64=0.

First determine the single solution in this cubic equation y3+64=0

Compare the equation y3+64=0 with the standard cubic equation ax3+bx2+cx+d=0.

a=1 and d=64

Therefore, possible factor of 1 is x=±1 and 64 is x=±1, x=±2, x=±4, x=±8, x=±16, x=±32 and x=±64.

First try y=4 substitute in the provided polynomial equation y3+64=0, (4)3+64=?0 64+64=?0         0=0

Which is true.

Therefore, y+4 is a factor of the polynomial y3+64=0.

In order to find other two values of x, divide y3+64=0 by y+4, y+4y24y+16y3+0y2+0y+64y3+4y2_   4y2+0y+64   4y216y_                16y +64&

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