Transcript Factoring Trinomials

ax2 + bx + c
Example 1:
x2 + 11x + 24
When factoring these trinomials the factors will be
two binomials: (x +
)(x +
)
We know that the first terms of each binomial must
be x because the first term of the trinomial is x2 and
x  x = x2. The challenge is to find the last term of
each binomial. They must be chosen so that they
will cause the coefficient of the middle term and
the last term of the trinomial to work out.
(That’s 11 and 24 in this case.)
Their sum equals the middle term (b) of the trinomial.
+
= 11
Their product of those same numbers equals the last
term © of the trinomial.
= 24

Example 1: x2 + 11x + 24
List the factors of 24:
24
1
24
SUM = 25
2
12
SUM = 14
3
8
SUM = 11
4
6
SUM = 10
11
It is the factors 3 and 8 that produce a sum of 11
AND a product of 24 so they must be the last terms
of each binomial.
(x + 3)(x + 8)
If we multiply these factors using FOIL, we get the
polynomial that we started with.
(x)(x) = x2
(x + 3)(x + 8)
= x2 + 8x + 3x + 24
(x)(8) = 8x
(3)(x) = 3x
(3)(8) = 24
As we look at the 4 terms above, it becomes apparent
why the sum of the last terms in each binomial must
be equal to the middle term of the trinomial.
(x + 3)(x + 8)
= x2 + 8x + 3x + 24
= x2 + 11x + 24
Example 2: a2 + 16a + 28
28
SUM = 16
16
a  a = a2 so they are the first terms of each binomial
and the factors 2 and 14 make a sum of 16 so the are
the last terms of each binomial.
= (a + 2)(a + 14)
Example 3: y2 + 2y + 1
y2
+ 2y + 1
= (y + 1)(y + 1)
Factors of 1:
1
1
SUM = 2
Sometimes there is only 1 pair of
factors to consider.
Factors of 1:
Example 4: m2 + 3m + 1
1
1
SUM = 3
In this example the factors available do
not make a sum of 3 which means that the Factors of 120:
trinomial can’t be factored.
1
120
Example 5: p2 + 23p + 120
In this example there are many pairs of
factors to consider. Most examples will
have fewer than these. The trick is in
being able to quickly find all of the factors
of c.
p2 + 23p + 120
= (p + 8)(p + 15)
2
60
3
40
4
30
5
24
6
20
8
15
10
12
SUM
= 23
Example 6: x2 + 5x + 6
= (x + 2)(x + 3)
Factors of 6:
1
6
2
3
SUM = 5
In each of the preceding examples the signs of the
terms in the trinomials were always positive. Now we
will observe examples where the signs can be negative.
Example 7: x2 + 5x - 6
= (x - 1)(x + 6)
Factors of -6:
-1
+6
-2
+3
SUM = 5
When looking for the factors of a negative number, one
must be positive and the other negative. If at the same
time their sum is positive, then the factor that is bigger
must be the positive one.
REVIEW OF RULES FOR SIGNS
ADDITION
MULTIPLICATION
(+)(+) = (+)
(+) + (+) = (+)
(+)(-) = (-)
(+) + (-) = Sign of bigger
number
(-)(+) = (-)
(-) + (+) =
(-)(-) = (+)
(-) + (-) = (-)
(
Example 8:
)
x2
- 5x - 6
= (x + 1)(x - 6)
When both the product and sum are
negative, the factors have opposite
signs but this time the bigger factor
will be negative.
-6
-5
Example 9: x2 - 5x + 6
= (x - 2)(x - 3)
Factors of 6:
-1
-6
-2
-3
SUM = -5
When looking for factors of a positive
number when the sum is negative,
both factors will be negative.
Example 10: x2 - 5x - 36
= (x + 4)(x - 9)
Factors of -36:
1
-36
2
-18
3
-12
4
-9
6
-6
SUM = -5
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 MAMA 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!!
Hang with me now! Replace the middle number of
the trinomial with our working numbers from the
MAMA 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?
There is a shortcut for some problems too!
(I’m not showing you that yet…)
M
A
2) Factor x2 – 2x – 63
Create your MAMA 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! Weeedoggie!
(x + 7)(x – 9)
Here are some hints to help
you choose your factors in the
MAMA 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 MAMA 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.
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 MAMA 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! Weeedoggie!
2(x – 6)(x – 1)
Don’t forget to follow your factoring chart when
doing these problems. Always look for a GCF
first!!