   Chapter A.4, Problem 25E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 25 to 32, solve each quadratic equation by factoring. x 2 - 6 x + 8 = 0

To determine

x2-6x+8=0

Explanation

Approach:

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,

x2-6x+8=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, -6x and 8.

 Terms Prime factors x2 x·x -6x 3·2·x·(-1) 8 2·2·2

But, the GCF of the terms x2, -6x and 8 is 1.

Hence, factor the expression x2-6x+8 by the reverse FOIL method.

First factorize the product of the coefficient of the first tern and the constant, such that the sum of the factors gives the coefficient of the middle term.

Thus, factorize 8 as -4·-2

So that the sum of the factors -4+-2 gives -6.

Now, split the middle term -6x as -4x+(-2x).

Hence, the given equation

Becomes.

x2+-4x+-2x+8=0

x·x+-4·x+-2·x+-2·-4=0

Factor out the common factors form the first two terms and form the last two terms of the above expression

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