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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Solving Equations by Factoring

Objective Learn to solve equations that are in factored form.

This lesson begins with a review of factoring, and then uses factored trinomials to solve equations. This technique (factoring) is very useful in problem solving.

## Factoring Expressions

The following are good examples.

Example 1

Factor x 2 - 4x - 21.

Solution

Since the coefficient of x 2 is 1, factor this expression as ( x + a )( x + b ) for some integer values of a and b. Now, ( x + a ) and ( x + b ) are multiplied to get x 2 + ( a + b )x + ab , which is supposed to equal x 2 - 4x - 21. Comparing x 2 + ( a + b )x + ab and x 2 - 4x - 21, we see that the sum a + b = -4 and the product ab = -21. So, the first step is to find the integral factors of - 21, for possible choices of a and b .

 - 21 = -21 Â· 1 - 21 = 21 Â· (-1) - 21 = -3 Â· 7 - 21 = 3 Â· (-7)

Look at the above factor pairs of -21 to see if any of them have a sum equal to - 4. If a = 3 and b = -7, then a + b = -4. The factorization is given below.

 x 2 - 4x - 21 = ( x + a )( x + b ) = ( x + 3 )[ x + (-7)] a = 3, b = -7 = ( x + 3 )( x - 7 )

You should multiply ( x + 3) and ( x - 7) to check that the factorization is correct.

Example 2

Factor 2x 3 - 3x 2 - 2x .

Solution

There is a common factor of x in each of the terms. So begin by factoring out x.

2x 3 - 3x 2 - 2x = x( 2x 2 - 3x - 2 )

The next step is to factor 2x 2 - 3x - 2. That is, write this expression as ( ax + b )( cx + d ) for some values of a, b, c, and d. When we multiply this out we get acx 2 + ( ad + bc )x + bd , which is supposed to equal 2x 2 - 3x - 2. So, ac = 2. Since 2 only has factors 2 and 1, let a = 2 and c = 1. We then have to find b and d. Since the coefficient of x , ad + bc , is now 2d + b Â· 1 or 2d + b , 2d + b = -3 and bd = -2. The factors of - 2 are 2 and 1, so b and d must be 2 or 1. On the other hand, 2d + b = -3. By checking the possibilities, we see that we must have d = -2 and b = 1. Substitute these values for a, b, c, and d.

 2x 3 - 3x 2 - 2x = ( ax + b )( cx + d ) = ( 2x + 1 )( x - 2 ) a = 2, b = 1, c = 1, d = -2

So,

 2x 3 - 3x 2 - 2x = x( 2x 2 - 3x - 2 ) = x( 2x + 1 )( x - 2 )

Again, you should multiply out x( 2x + 1 )( x - 2 ) to check the factoring.