Transcript Factor Trinomials by Grouping

Objective
The student will be able to:
factor trinomials with grouping
2
Quadratic Equations: x + bx + c = 0
Factoring Chart
This chart will help you to determine
which method of factoring to use.
Type
Number of Terms
1. GCF
2 or more
2. Diff. Of Squares 2
3. Trinomials
3
Review: (y + 2)(y + 4)
y2
+4y
+2y
+8
First terms:
Outer terms:
Inner terms:
Last terms:
Combine like terms.
y2 + 6y + 8
y
+2
y2
+2y
+4 +4y
+8
y
In this lesson, we will begin with y2 + 6y + 8 as our
problem and finish with (y + 2)(y + 4) as our answer.
Here we go! 1) Factor y2 + 6y + 8
Use your factoring chart.
Do we have a GCF? Nope!
Is it a Diff. of Squares problem? No way! 3 terms!
Now we will learn Trinomials! You will set up
a table with the following information.
Product of the first and
last coefficients
Middle
coefficient
The goal is to find two factors in the first column that
add up to the middle term in the second column.
We’ll work it out in the next few slides.
1) Factor
2
y
M
A
+ 6y + 8
Create your table.
Product of the
first and last
coefficients
Multiply
+8
Add
+6
Middle
coefficient
Here’s your task…
What numbers multiply to +8 and add to +6?
If you cannot figure it out right away, write
the combinations.
1) Factor
2
y
+ 6y + 8
Place the factors in the table.
Multiply
+8
Which has
a sum
of +6?
+1, +8
-1, -8
+2, +4
-2, -4
Add
+6
+9, NO
-9, NO
+6, YES!!
-6, NO
We are going to use these numbers in the next step!
1) Factor y2 + 6y + 8
Multiply
+8
Add
+6
+2, +4 +6, YES!!
Replace the middle number of the trinomial with
our working numbers from the MA table
y2 + 6y + 8
y2 + 2y + 4y + 8
Now, group the first two terms and the last two
terms.
We have two groups!
(y2 + 2y)(+4y + 8)
Almost done! Find the GCF of each group and factor
it out.
If things are done
right, the parentheses
y(y + 2) +4(y + 2)
should be the same.
Factor out the
GCF’s. Write them
in their own group.
(y + 4)(y + 2)
Tadaaa! There’s your answer…(y + 4)(y + 2)
You can check it by multiplying. Piece of cake, huh?
M
A
2) Factor x2 – 2x – 63
Create your table.
Product of the
first and last
coefficients
Signs need to
be different
since number
is negative.
Multiply
-63
-63, 1
-1, 63
-21, 3
-3, 21
-9, 7
-7, 9
Add
-2
-62
62
-18
18
-2
2
Middle
coefficient
Replace the middle term with our working
numbers.
x2 – 2x – 63
x2 – 9x + 7x – 63
Group the terms.
(x2 – 9x) (+ 7x – 63)
Factor out the GCF
x(x – 9) +7(x – 9)
The parentheses are the same!
(x + 7)(x – 9)
Here are some hints to help
you choose your factors in the
table.
1) When the last term is positive, the factors
will have the same sign as the middle term.
2) When the last term is negative, the factors
will have different signs.
M
A
2) Factor 5x2 - 17x + 14
Create your table.
Product of the
first and last
coefficients
Signs need to
be the same as
the middle
sign since the
product is
positive.
Multiply
+70
-1, -70
-2, -35
-7, -10
Add
-17
-71
-37
-17
Replace the middle term.
5x2 – 7x – 10x + 14
Group the terms.
Middle
coefficient
(5x2 – 7x) (– 10x + 14)
Factor out the GCF
x(5x – 7) -2(5x – 7)
The parentheses are the same!
(x – 2)(5x – 7)
Hopefully, these will continue to get easier the
more you do them.
Factor
1.
2.
3.
4.
(x + 2)(x + 1)
(x – 2)(x + 1)
(x + 2)(x – 1)
(x – 2)(x – 1)
2
x
+ 3x + 2
Factor
1.
2.
3.
4.
(2x + 10)(x + 1)
(2x + 5)(x + 2)
(2x + 2)(x + 5)
(2x + 1)(x + 10)
2
2x
+ 9x + 10
Factor
1.
2.
3.
4.
2
6y
(6y2 – 15y)(+2y – 5)
(2y – 1)(3y – 5)
(2y + 1)(3y – 5)
(2y – 5)(3y + 1)
– 13y – 5
2) Factor 2x2 - 14x + 12
Find the GCF!
2(x2 – 7x + 6)
Now do the MA table!
Signs need to
be the same as
the middle
sign since the
product is
positive.
Multiply
+6
Add
-7
-1, -6
-7
-2, -3
-5
Replace the middle term.
2[x2 – x – 6x + 6]
Group the terms.
2[(x2 – x)(– 6x + 6)]
Factor out the GCF
2[x(x – 1) -6(x – 1)]
The parentheses are the same!
2(x – 6)(x – 1)
Don’t forget to follow your factoring chart when
doing these problems. Always look for a GCF
first!!