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Algebra 1

6.5.4 Factoring Trinomials Using the “ac” Method and Substitution

Algebra 16.5.4 Factoring Trinomials Using the “ac” Method and Substitution

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Activity

Let’s learn a new method that always works on any trinomial of the form a x 2 + b x + c a x 2 + b x + c .

How to factor trinomials of the form a x 2 + b x + c a x 2 + b x + c using the “ a c a c ” method:

Step 1 - Factor any GCF. Don’t forget!

Step 2 - Find the product a c a c .

Step 3 - Find two numbers, m m and n n , that

  • multiply to a c a c : m · n = a · c m · n = a · c .
  • add to b b : m + n = b m + n = b .

Step 4 - Split the middle term using m m and n n .

a x 2 + m x + n x + c a x 2 + m x + n x + c

Step 5 - Factor by grouping.

Step 6 - Check by multiplying the factors.

Factor each trinomial using the “ a c a c ” method.

1. 16 x 2 32 x + 12 16 x 2 32 x + 12

2. 6 y 2 + y 15 6 y 2 + y 15

3. 48 z 3 102 z 2 45 z 48 z 3 102 z 2 45 z

There will be times when a trinomial does not seem to be in the form a x 2 + b x + c a x 2 + b x + c .

In some of these cases, we can make a substitution that allows the trinomial to be written in the a x 2 + b x + c a x 2 + b x + c form. This is called factoring by substitution.

It is standard to use u u for the substitution. The substitution for u u might be a variable with an exponent, such as x 2 x 2 , or it might be a binomial, such as ( x 3 ) ( x 3 ) .

Factor each trinomial using the substitution method.

4. y 4 y 2 20 y 4 y 2 20

5. ( x 5 ) 2 + 6 ( x 5 ) + 8 ( x 5 ) 2 + 6 ( x 5 ) + 8

6. x 4 3 x 2 28 x 4 3 x 2 28

Self Check

Factor using the “ a c ” method: 6 x 2 + 13 x + 2 .
  1. ( x 2 ) ( 6 x + 1 )
  2. ( x 2 ) ( 6 x 1 )
  3. ( 6 x + 2 ) ( x + 1 )
  4. ( x + 2 ) ( 6 x + 1 )

Additional Resources

Factoring Trinomials Using the “ a c a c ” Method

Another way to factor trinomials of the form a x 2 + b x + c a x 2 + b x + c is the “ a c a c ” method. (The “ a c a c ” method is sometimes called the grouping method.) The “ a c a c ” method is actually an extension of the methods you used in the last section to factor trinomials with a leading coefficient of 1. This method is very structured (that is, step by step), and it always works!

Example 1

Factor using the “ a c a c ” method: 6 x 2 + 7 x + 2 6 x 2 + 7 x + 2 .

Step 1 - Factor any GCF.

Is there a greatest common factor?

6 x 2 + 7 x + 2 6 x 2 + 7 x + 2

No!

Step 2 - Find the product a c a c .

a x 2 + b x + c 6 x 2 + 7 x + 2 a x 2 + b x + c 6 x 2 + 7 x + 2

6 · 2 6 · 2

12 12

Step 3 - Find two numbers, m m and n n , that

  • multiply to a c a c : m · n = a · c m · n = a · c .
  • add to b b : m + n = b m + n = b .

Find two numbers that multiply to 12 and add to 7. Both factors must be positive.

3 · 4 = 12 3 · 4 = 12

3 + 4 = 7 3 + 4 = 7

Step 4 - Split the middle term using m m and n n .

Mathematical expression showing ax squared  + bx + c, with bx split as mx + nx, and a bracket indicating bx is replaced by mx + nx in ax squared + bx + c.

Rewrite 7 x 7 x as 3 x + 4 x 3 x + 4 x . It would give the same result if we used 4 x + 3 x 4 x + 3 x .

The quadratic expression 6x squared + 7x + 2 is split into 6x squared  + 3x and 4x + 2, with arrows pointing from 7x to 3x and 4x, showing the process of factoring by grouping.

Notice that 6 x 2 + 3 x + 4 x + 2 6 x 2 + 3 x + 4 x + 2 is equal to 6 x 2 + 7 x + 2 6 x 2 + 7 x + 2 . We just split the middle term to get a more useful form.

Step 5 - Factor by grouping.

3 x ( 2 x + 1 ) + 2 ( 2 x + 1 ) 3 x ( 2 x + 1 ) + 2 ( 2 x + 1 )

( 2 x + 1 ) ( 3 x + 2 ) ( 2 x + 1 ) ( 3 x + 2 )

Step 6 - Check by multiplying the factors.

( 2 x + 1 ) ( 3 x + 2 ) ( 2 x + 1 ) ( 3 x + 2 )

6 x 2 + 4 x + 3 x + 2 6 x 2 + 4 x + 3 x + 2

6 x 2 + 7 x + 2 6 x 2 + 7 x + 2

The “ a c a c ” method is summarized here:

How to factor trinomials of the form a x 2 + b x + c a x 2 + b x + c using the “ a c a c ” method:

Step 1 - Factor any GCF. Don’t forget!

Step 2 - Find the product a c a c .

Step 3 - Find two numbers, m m and n n , that

  • multiply to a c a c : m · n = a · c m · n = a · c .
  • add to b b : m + n = b m + n = b .

Step 4 - Split the middle term using m m and n n .

a x 2 + m x + n x + c a x 2 + m x + n x + c

Step 5 - Factor by grouping.

Step 6 - Check by multiplying the factors.

Try it

Try It: Factoring Trinomials Using the “ a c a c ” Method

Factor using the “ a c a c ” method: 10 y 2 55 y + 70 10 y 2 55 y + 70 .

Factoring Trinomials Using Substitution

Sometimes a trinomial does not appear to be in the a x 2 + b x + c a x 2 + b x + c form. However, we can often make a thoughtful substitution that will allow us to make it fit the a x 2 + b x + c a x 2 + b x + c form. This is called factoring by substitution. It is standard to use u u for the substitution.

In the form a x 2 + b x + c a x 2 + b x + c , the middle term has a variable, x x , and its square, x 2 x 2 , is the variable part of the first term. Look for this relationship as you try to find a substitution.

Example 2

Factor by substitution: x 4 4 x 2 5 x 4 4 x 2 5 .

The variable part of the middle term is x 2 x 2 , and its square, x 4 x 4 , is the variable part of the first term. (We know ( x 2 ) 2 = x 4 ) ( x 2 ) 2 = x 4 ) . If we let u = x 2 u = x 2 , we can put our trinomial in the a x 2 + b x + c a x 2 + b x + c form we need to factor it.

Step 1 - Rewrite the trinomial to prepare for the substitution.

( x 2 ) 2 4 ( x 2 ) 5 ( x 2 ) 2 4 ( x 2 ) 5

Step 2 - Let u = x 2 u = x 2 and substitute.

u 2 4 u 5 u 2 4 u 5

Step 3 - Factor the trinomial.

( u + 1 ) ( u 5 ) ( u + 1 ) ( u 5 )

Step 4 - Replace u u with x 2 x 2 .

( x 2 + 1 ) ( x 2 5 ) ( x 2 + 1 ) ( x 2 5 )

Step 5 - Check:

( x 2 + 1 ) ( x 2 5 ) x 4 5 x 2 + x 2 5 x 4 4 x 2 5 ✓ ( x 2 + 1 ) ( x 2 5 ) x 4 5 x 2 + x 2 5 x 4 4 x 2 5 ✓ ( x 2 + 1 ) ( x 2 5 ) x 4 5 x 2 + x 2 5 x 4 4 x 2 5 ✓

Try it

Try It: Factoring Trinomials Using Substitution

Factor by substitution: ( x 2 ) 2 + 7 ( x 2 ) + 12 ( x 2 ) 2 + 7 ( x 2 ) + 12 .

In this case, the expression to be substituted is not a monomial. Let u = x 2 u = x 2 .

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