   Chapter 0.5, Problem 43E

Chapter
Section
Textbook Problem

Find all possible real solutions of each equation in Exercises 31–44. ( x 2 + 3 x + 2 ) ( x 2 − 5 x + 6 ) = 0

To determine

To calculate: The all real solutions in the polynomial equation (x2+3x+2)(x25x+6)=0.

Explanation

Given Information:

The provided polynomial equation is (x2+3x+2)(x25x+6)=0.

Formula used:

Principle of 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±b4ac2a where a,b and c are real numbers.

Calculation:

Consider the polynomial equation (x2+3x+2)(x25x+6)=0.

Apply the zero-product rule,

If x2+3x+2=0

Compare the above equation with the equation ax2+bx+c=0,

a=1

, b=3 and c=2

Put the values of a,b,c in the quadratic formula x=b±b24ac2a.

x=3±(3)241221=3±982=3±12=3±12

Further simplify,

x=

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