Transcript Unit 10-2 Objectives The student will be able to:

Objective
The student will be able to:
use grouping to factor
polynomials with four terms.
SOL: A.2c
Designed by Skip Tyler, Varina High School
1. Factor rx + 2ry + kx + 2ky
You have 4 terms - try factoring by grouping.
(rx + 2ry) + (kx + 2ky)
Find the GCF of each group.
r(x + 2y) + k(x + 2y)
The parentheses are the same!
(r + k)(x + 2y)
2. Factor
2
2x
- 3xz - 2xy + 3yz
Check for a GCF: None
Factor by grouping. Keep a + between the groups.
(2x2 - 3xz) + (- 2xy + 3yz)
Find the GCF of each group.
x(2x – 3z) + y(- 2x + 3z)
The signs are opposite in the parentheses!
Keep-change-change!
x(2x - 3x) - y(2x - 3z)
(x - y)(2x - 3z)
Objective
The student will be able to:
factor trinomials with grouping.
SOL: A.2c
Designed by Skip Tyler, Varina High School
Here we go! 1) Factor x2 + 3x + 2
Use your factoring chart.
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
x
M
A
+ 3x + 2
Create your MA table.
Product of the
first and last
coefficients
1st coefficient is an
unwritten “1” ,
the last coefficient
is 2
Multiply
+2
Add
+3
Middle
coefficient
Here’s your task…
What numbers multiply to +2 and add to +3?
If you cannot figure it out right away, write
the combinations.
1) Factor
2
x
+ 3x + 2
Place the factors in the table.
Only 2
Factors
are 2 and
1.
Multiply
+2
Add
+3
+2, +1 +3, YES!!
We are going to use these numbers in the next step!
1) Factor x2 + 3x + 2
Multiply
+2
Add
+3
+2, +1 +3, YES!!
Hang with me now! Replace the middle number of
the trinomial with our working numbers from the
MA table
x2 + 3x + 2
x2 + 2x + 1x + 2
Now, group the first two terms and the last two
terms.
We have two groups!
(x2 + 2x)(+1x + 2)
Almost done! Find the GCF of each group and factor
it out.
If things are done
right, the parentheses
x(x + 2) +1(x + 2)
should be the same.
Factor out the
GCF’s. Write them
in their own group.
(x + 1)(x + 2)
Tadaaa! There’s your answer…(x + 1)(x + 2)
You can check it by multiplying. Piece of cake, huh?
There is a shortcut for some problems too!
(I’m not showing you that yet…)
M
A
2) Factor 5x2 - 17x + 14
Create your MA 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! Weeedoggie!
(x – 2)(5x – 7)
Hopefully, these will continue to get easier the
more you do them.
Here are some hints to help
you choose your factors in the
MA 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.
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