Module 21.2

Solving EquationsBy Factoring 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄

How can you use factoring to solve quadratic equations in standard form for which a ≠ 1?

P. 997

Until now we’ve been factoring quadratic expressions where the leading coefficient “a” has been 1.

For example:

What do we do when the leading coefficient is NOT equal 1?

For example:

We can’t use the standard form of 𝑥 + ? 𝑥 + ? because of the 𝟒.

There are a few different methods that are used to factor these, such as Slide & Divide, Guess & Check, Tic-Tac-Toe, and Grouping.We’re going to learn the Box method.

𝒙𝟐 + 𝟔𝒙 + 𝟖

𝟒𝒙𝟐 + 𝟏𝟎𝒙 + 𝟔

𝟒𝒙𝟐 + 𝟏𝟎𝒙 + 𝟔

BOX METHOD - 9 Steps

1) Determine if there’s a GCF for all 3 terms. If yes, then factor it out.Here, the GCF is 2, so it becomes:

2) Before: Find all factors that multiply to c and add up to bNow: Find all factors that multiply to a·c and add up to b.Here: Find all factors that multiply to 6 and add up to 5.

1,62,3 <<== Which gives us 2x and 3x

𝟐(𝟐𝒙𝟐 + 𝟓𝒙 + 𝟑)

a b c

3) Create a 2x2 table.In the upper left, put the first term.In the lower right, put the last term.In the other two, put the 2 new terms from the previous step. (It doesn’t mater which of those 2 goes where.)

a b c

𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

𝟐(𝟐𝒙𝟐 + 𝟓𝒙 + 𝟑)

4) Between the top 2 boxes, determine the GCF.Write that to the left.

2x 𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

5) Divide the upper-left box by the number you just wrote,and write that new number on top of the upper-left box.

2x 𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

x

𝟐(𝟐𝒙𝟐 + 𝟓𝒙 + 𝟑)

6) Divide the upper-right box by the number you wrote to the left,and write that new number on top of the upper-right box.

2x 𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

x 1

7) Divide the lower-left box by the number you wrote at the top,and write that new number to the left of the lower-left box.You should now have what looks like a multiplication table.

2x3

𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

x 1

𝟐(𝟐𝒙𝟐 + 𝟓𝒙 + 𝟑)

8) Use the numbers you’ve written to create two binomials,and combine it with the GCF from the first step, if any.

2x3

𝟐𝒙𝟐 𝟐𝒙

𝟑𝒙 𝟑

x 1

𝟐 𝒙 + 𝟏 𝟐𝒙 + 𝟑

9) Check your work by multiplying this out (via FOIL).Does it equal the original expression?

𝟐 𝒙 + 𝟏 𝟐𝒙 + 𝟑 = 𝟐 𝟐𝒙𝟐 + 𝟑𝒙 + 𝟐𝒙 + 𝟑

= 𝟐 𝟐𝒙𝟐 + 𝟓𝒙 + 𝟑

= 𝟒𝒙𝟐 + 𝟏𝟎𝒙 + 𝟔

Yes!

𝟔𝒙𝟐 + 𝟏𝟗𝒙 + 𝟏𝟎

Let’s try another one.

1) Is there a GCF? No.2) Find all factors that multiply to 60 and add up to 19.

1,602,303,204,15 <<== Which gives us 4x and 15x

3) Create a 2x2 table with the 4 terms.

𝟔𝒙𝟐 𝟒𝒙

𝟏𝟓𝒙 𝟏𝟎

4) Between the top 2 boxes, determine the GCF, and write that to the left.5) Divide the upper-left box by the number you just wrote,

and write that new number on top of the upper-left box.6) Divide the upper-right box by the number you wrote to the left,

and write that new number on top of the upper-right box.7) Divide the lower-left box by the number you wrote at the top,

and write that new number to the left of the lower-left box.8) Use the numbers you’ve written to create two binomials,

and combine it with the GCF from the first step, if any.9) Check your work by multiplying this out (via FOIL).

2x5

3x 2

𝟑𝒙 + 𝟐 𝟐𝒙 + 𝟓𝟔𝒙𝟐 𝟒𝒙

𝟏𝟓𝒙 𝟏𝟎

Practice:

𝟔𝒙𝟐 − 𝟐𝟏𝒙 − 𝟒𝟓

𝟐𝒙𝟐 − 𝟕𝒙 + 𝟔

P. 999-1000

−𝟓𝒙𝟐 + 𝟖𝒙 + 𝟒

P. 1000

1st # 2nd # Added

1 12 13

2 6 8

3 4 7

–1 –12 –13

–2 –6 –8

–3 –4 –7

P. 1000

These are the Solutions aka X-interceptsaka Zeros aka Roots

Use the Zero Product Property

P. 1001

P. 1001

P. 1002

P. 1002

P. 1003

P. 1003