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Algebra 1

8.10.2 Working with Quadratic Factored Form with Leading Coefficient Other Than One

Algebra 18.10.2 Working with Quadratic Factored Form with Leading Coefficient Other Than One

8.10.2 Working with Quadratic Factored Form with Leading Coefficient Other Than One

8.10.2 • Working with Quadratic Factored Form with Leading Coefficient Other Than One

Activity

Each row in the given tables contains a pair of equivalent expressions. Complete the tables by converting the given expression into standard form. If you get stuck, try drawing a diagram.

1.

Factored form Standard form

( 3 x + 1 ) ( x + 4 ) ( 3 x + 1 ) ( x + 4 )

 

2.

Factored form Standard form

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

 

3.

Factored form Standard form

( 3 x + 4 ) ( x + 1 ) ( 3 x + 4 ) ( x + 1 )

 

4. To find the factored form of a quadratic expression with a leading coefficient other than 1, answer parts a–d.

Factored form Standard form
  5 x 2 + 21 x + 4 5 x 2 + 21 x + 4

a. This quadratic expression has a leading coefficient of 5. Identify the factors of 5 to determine the leading term of each binomial in factored form.

( _ _ _ x + _ _ _ ) ( _ _ _ x + _ _ _ ) ( _ _ _ x + _ _ _ ) ( _ _ _ x + _ _ _ )

b. The constant term in this quadratic expression is +4. Identify all the factors of +4.

c. The linear term in the expression is 21 x 21 x . Place each of the factors of +4 you found in part b into the remaining blanks of the factored form structure from part a. Which of these results in a linear term of 21 x 21 x when you use FOIL? (Keep in mind that the order you place these factors into your factored form structure might affect the outcome, so try both orders if needed.)

d. What is the factored form of 5 x 2 + 21 x + 4 5 x 2 + 21 x + 4 ?

5. Use the strategy you developed in question 4 to find the factored form of the quadratic expression shown here in standard form.

Factored form Standard form
  3 x 2 + 7 x + 4 3 x 2 + 7 x + 4

Are you ready for more?

Extending Your Thinking
1.

Find the factored form of the quadratic expression shown.

Factored form Standard form
 

6 x 2 + 11 x − 10 6 x 2 + 11 x − 10

This quadratic expression is different from those in the activity, since its leading coefficient has more than one set of possible factors.

Use the strategy you learned in the activity to factor the expression. If the pair of factors you selected using the leading coefficient, 6, do not give you the correct solution, begin again with another set of factors.

The expression is also different since the constant term is negative, -10. This means when you consider the factors of the constant term, one must be positive and one must be negative.

Self Check

Find the factored form of the quadratic expression 3 x 2 + 5 x − 2 .
  1. ( 3 x − 1 ) ( x + 2 )
  2. ( 3 x − 2 ) ( x + 1 )
  3. ( 3 x + 2 ) ( x + 1 )
  4. ( 3 x − 2 ) ( x − 1 )

Additional Resources

Working with Expressions of the Form ( p x + m ) ( q x + n ) ( p x + m ) ( q x + n )

You have factored and solved many different types of quadratic equations, but the leading coefficient of the x 2 x 2 term was often 1.

Let’s look at the factored form of a more complex expression to better understand the process.

Example 1

Write ( 4 x + 1 ) ( x + 3 ) ( 4 x + 1 ) ( x + 3 ) in standard form.

To find the quadratic expression in standard form, use the distributive property of multiplication over addition, or FOIL.

( 4 x + 1 ) ( x + 3 ) = 4 x 2 + 12 x + x + 3 = 4 x 2 + 13 x + 3 ( 4 x + 1 ) ( x + 3 ) = 4 x 2 + 12 x + x + 3 = 4 x 2 + 13 x + 3

Now, let’s try to go the other direction from standard form to factored form.

Example 2

Write 2 x 2 + 13 x + 15 2 x 2 + 13 x + 15 in factored form.

Since the leading coefficient is 2, we know the factored form must have the structure:

( 2 x + _ _ _ ) ( x + _ _ _ ) ( 2 x + _ _ _ ) ( x + _ _ _ )

Since the constant term is +15, the two missing numbers in the factored form must have a product of 15.

The possibilities include:

+1 and +15

+3 and +5

-1 and -15

-3 and -5

Since the middle term, 13 x 13 x , is positive, we know that only the positive factors need to be considered.

Since the two factors are already different inside the parentheses, the order in which we place the numbers is important. We will try both, if necessary.

Let’s try 1 1 and 1 5 1 5 .

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

= 2 x 2 + 2 x + 15 x + 15 = 2 x 2 + 17 x + 15 = 2 x 2 + 2 x + 15 x + 15 = 2 x 2 + 17 x + 15

not equivalent to the original expression

Try reordering the same numbers.

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

= 2 x 2 + 30 x + x + 15 = 2 x 2 + 31 x + 15 = 2 x 2 + 30 x + x + 15 = 2 x 2 + 31 x + 15

not equivalent to the original expression

Let’s try 3 3 and 5 5 .

( 2 x + 5 ) ( x + 3 ) ( 2 x + 5 ) ( x + 3 )

= 2 x 2 + 6 x + 5 x + 15 = 2 x 2 + 11 x + 15 = 2 x 2 + 6 x + 5 x + 15 = 2 x 2 + 11 x + 15

not equivalent to the original expression

Try reordering the same numbers.

( 2 x + 3 ) ( x + 5 ) ( 2 x + 3 ) ( x + 5 )

= 2 x 2 + 10 x + 3 x + 15 = 2 x 2 + 13 x + 15 = 2 x 2 + 10 x + 3 x + 15 = 2 x 2 + 13 x + 15

✓ ✓ equivalent to the original expression

So, the factored form of 2 x 2 + 13 x + 15 2 x 2 + 13 x + 15 is ( 2 x + 3 ) ( x + 5 ) ( 2 x + 3 ) ( x + 5 ) .

Try it

Working with Expressions of the Form ( p x + m ) ( q x + n ) ( p x + m ) ( q x + n )

Find the factored form of 3 x 2 − 11 x − 4 3 x 2 − 11 x − 4 .

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