Chapter A.4, Problem 40E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# With the highest degree term being 4 x 3 , the equation 4 x 3 - 28 x 2 + 24 x = 0 is known as a cubic equation. Use the factoring to find three real solutions for this equation.

To determine

The cubic equation 4x3-28x2+24x=0, by using factorization.

Explanation

Solving an equation is to find the value of the unknown variables in the equation, such that the obtained value or values of the unknown should satisfy the equation from which it was derived. Such a value is said to be the solution for the equation. In general a quadratic equation has two solutions for the variable in the equation as the degree of the equation is two.

Calculation:

Given,

4x3-28x2+24x=0

Now, we shall proceed with the factorization of the quadratic expression in the left of the equation.

First look for the common factor GCF of the terms 4x3, -28x2 and 24x.

 Terms Prime factors 4x3âˆ’28x224x 2â‹…2â‹…xâ‹…xâ‹…x2â‹…2â‹…7â‹…xâ‹…xâ‹…(âˆ’1)2â‹…â€‰2â‹…2â‹…3â‹…x

Thus, the GCF is 2Â·2Â·x=4x.

The given equation become,

4xÂ·x2-4xÂ·7x+4xÂ·6=0

Since, 4x is distributed over the expression x2-7x+6, we have

4xÂ·x2-7x+6=0

4xx2-7x+6=0

Divide; by 4 on both side,

4xx2-7x+64=04

xx2-7x+6=0

Now, by the zero product property

x=0; x2-7x+6=0

Take;

x2-7x+6=0

The above equation is in the standard form of a quadratic equation.

Now, we shall proceed with the factorization of the quadratic expression in the left of the equation.

First look for the common factor GCF of the terms x2, -7x and 6.

But, here the GCF is 1

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