Download Chapter 4: Factoring Polynomials

Topic:
Factoring x2 + bx + c
Essential Question:
How can you write a trinomial as the product of two binomials?

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Home-Learning Review
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Quiz #10
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Topic: Factoring
2
x + bx + c
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Factors
Factors (either numbers or polynomials)
When an integer is written as a product of
integers, each of the integers in the product is a
factor of the original number.
When a polynomial is written as a product of
polynomials, each of the polynomials in the
product is a factor of the original polynomial.
Factoring – writing a polynomial as a product of
polynomials.
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Greatest Common Factor
Greatest common factor – largest quantity that is a
factor of all the integers or polynomials involved.
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Greatest Common Factor
Example
Find the GCF of each list of numbers.
A) 12 and 8
12 =
8=
So the GCF is
B) 6, 8, and 46
6=
8=
46 =
So the GCF is
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Greatest Common Factor
Example
Find the GCF of each list of terms.
C) x3 and x7
x3 = x · x · x
x7 = x · x · x · x · x · x · x
So the GCF is x · x · x = x3
D) 6x5 and 4x3
6x5 = 2 · 3 · x · x · x· x · x
4x3 = 2 · 2 · x · x · x
So the GCF is 2 · x · x · x = 2x3
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Greatest Common Factor
Example
Find the GCF of the following list of terms.
E) a3b2, a2b5 and a4b7
a3b2 = a · a · a · b · b
a2b5 = a · a · b · b · b · b · b
a4b7 = a · a · a · a · b · b · b · b · b · b · b
So the GCF is a · a · b · b = a2b2
Notice that the GCF of terms containing variables will use the
smallest exponent found amongst the individual terms for each
variable.
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Lets make the connection!
Ready
Set
Go
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Factoring Polynomials
The first step in factoring a polynomial is to
find the GCF of all its terms.
Then we write the polynomial as a product by
factoring out the GCF from all the terms.
The remaining factors in each term will form a
polynomial.
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Factoring out the GCF
Example
Factor out the GCF in each of the following
polynomials.
F) 6x3 – 9x2 + 12x =
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Factoring out the GCF
Practice
Factor out the GCF in each of the following
polynomials.
3) 14x3y + 7x2y – 7xy =
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Factoring Trinomials of the
2
Form x + bx + c
Factoring Trinomials
Recall by using the FOIL method that
F
O
I
L
(x + 2)(x + 4) = x2 + 4x + 2x + 8
= x2 + 6x + 8
To factor x2 + bx + c into (x + one #)(x + another #),
note that b is the sum of the two numbers and c is the
product of the two numbers.
So we’ll be looking for 2 numbers whose product is
c and whose sum is b.
Note: there are fewer choices for the product, so
that’s why we start there first.
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Factoring Polynomials
Example
G) Factor the polynomial x2 + 13x + 30.
Since our two numbers must have a product of 30 and a
sum of 13, the two numbers must both be positive.
Positive factors of 30
Sum of Factors = 13
1 x 30
1 + 30 = 31 x
2 x 15
2 + 15 = 17 x
3 x 10
3 + 10 = 13 √
Note, there are other factors, but once we find a pair
that works, we do not have to continue searching.
So x2 + 13x + 30 = (x + 3)(x + 10).
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x  6x  8
2
Example :
x
x
-2
 x
x
Factors of +8:
-4


x x  x
1&8
1x + 8x = 9x
2&4
2x + 4x = 6x
-1 & -8
-1x - 8x = -9x
-2 & -4
-2x - 4x = -6x
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x  6x  8  (x  2)(x  4)
2
Check your answer by
using FOIL
F
O
I
L
2
(x  2)(x  4)  x  4x  2x  8
2
 x  6x  8
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Factoring Polynomials
Practice:
4) Factor the polynomial x2 – 11x + 24.
Since our two numbers must have a product of 24 and a
sum of -11, the two numbers must both be negative.
Negative factors of 24
Sum of Factors = –11
– 1 x – 24
– 1 + – 24= – 25 x
– 2 x – 12
– 2 + – 12 = – 14 x
–3x–8
– 3 + – 8 = – 11 √
So x2 – 11x + 24 = (x – 3)(x – 8).
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Factoring Polynomials
Practice:
5) Factor the polynomial x2 – 2x – 35
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Prime Polynomials
Example
Factor the polynomial x2 – 6x + 10.
Since our two numbers must have a product of 10 and a
sum of – 6, the two numbers will have to both be negative.
Negative factors of 10
Sum of Factors
– 1, – 10
– 11
– 2, – 5
–7
Since there is not a factor pair whose sum is – 6,
x2 – 6x +10 is not factorable and we call it a prime
polynomial.
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Check Your Result!
You should always check your factoring
results by multiplying the factored polynomial
to verify that it is equal to the original
polynomial.
Many times you can detect computational
errors or errors in the signs of your numbers
by checking your results.
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Pair Practice:
Page 483
(16, 22)
Page 503 - 505
(10, 12, 14, 16, 20, 22, 24,
32, 42, 61, 62)
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Additional Practice
Warm-Up:

– 36p
2
 m + 8m
2
 x + 2x – 8
2
16p
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Factoring
Trinomials of
the Form
2
ax + bx + c
Working with a value of ‘a’ that is higher than 1
Factoring Trinomials
Returning to the FOIL method,
F
O
I L
(3x + 2)(x + 4) = 3x2 + 12x + 2x + 8
= 3x2 + 14x + 8
To factor ax2 + bx + c into (#1·x + #2)(#3·x + #4),
note that a is the product of the two first coefficients,
c is the product of the two last coefficients and b is
the sum of the products of the outside coefficients
and inside coefficients.
Note that b is the sum of 2 products, not just 2
numbers, as in the last section.
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Factoring Polynomials
Example
A)
Factor the polynomial 25x2 + 20x + 4
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Factoring Polynomials
Example Continued
Check the resulting factorization using the FOIL method.
F
O
I
L
(5x + 2)(5x + 2) = 5x(5x) + 5x(2) + 2(5x) + 2(2)
= 25x2 + 10x + 10x + 4
= 25x2 + 20x + 4
So our final answer when asked to factor 25x2 + 20x + 4
will be (5x + 2)(5x + 2) or (5x + 2)2.
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Factoring Polynomials
Example
B) Factor the polynomial 21x2 – 41x + 10.
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Factoring Polynomials
Example Continued
Check the resulting factorization using the FOIL method.
F
O
I
L
(3x – 5)(7x – 2) = 3x(7x) + 3x(-2) - 5(7x) - 5(-2)
= 21x2 – 6x – 35x + 10
= 21x2 – 41x + 10
So our final answer when asked to factor 21x2 – 41x + 10
will be (3x – 5)(7x – 2).
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Factoring Trinomials (Method 2*)
This time, the x2
term DOES
have a
coefficient
(other than 1)!
Factor. 3x2 + 14x + 8
Step 1: Multiply 3 • 8 = 24
(the leading coefficient & constant).
Step 2: List all pairs of
numbers that multiply to equal
that product, 24.
1 • 24 = 24
2 • 12 = 24
3 • 8 = 24
4 • 6 = 24
Step 3: Which pair adds up to 14?
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Factoring Trinomials (Method 2*)
Factor. 3x2 + 14x + 8
Step 4: Write temporary
factors with the two numbers.
( x + 2 )( x + 12 )
Step 5: Put the original
leading coefficient (3) under
both numbers.
( x + 2 )( x + 12 )
3
3
Step 6: Reduce the fractions, if
possible.
( x + 2 )( x + 12 )
3
3
Step 7: Move denominators in
front of x.
( 3x + 2 )( x + 4 )
4
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Factoring Trinomials (Method 2*)
Factor.
3x2 + 14x + 8
You should always check the factors by distributing, especially
since this process has more than a couple of steps.
( 3x + 2 )( x + 4 ) = 3x • x + 3x • 4 + 2 • x + 2 • 4
= 3x2 + 14 x + 8
√
3x2 + 14x + 8 = (3x + 2)(x + 4)
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Pair-Practice:
Practice:
1)
3x2 + 13x + 4
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Pair-Practice:
Practice:
2)
2x2 + x – 21
3)
25x2 + 20x + 4
4)
3r2 – 18r + 27
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Factoring Polynomials
Example
C)
Factor the polynomial 3x2 – 7x + 6.
The only possible factors for 3 are 1 and 3, so we know that, if
factorable, the polynomial will have to look like (3x
)(x
)
in factored form, so that the product of the first two terms in the
binomials will be 3x2.
Since the middle term is negative, possible factors of 6 must both
be negative: {1,  6} or { 2,  3}.
We need to methodically try each pair of factors until we find a
combination that works, or exhaust all of our possible pairs of
factors.
Continued.
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Factoring Polynomials
Example Continued
We will be looking for a combination that gives the sum of the
products of the outside terms and the inside terms equal to 7x.
Now we have a problem, because we have
exhausted all possible choices for the factors,
but have not found a pair where the sum of the
products of the outside terms and the inside
terms is –7.
So 3x2 – 7x + 6 is a prime polynomial and will
not factor.
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Factoring Polynomials
Example
D)
Factor the polynomial 6x2y2 – 2xy2 – 60y2.
Remember that the larger the coefficient, the greater the
probability of having multiple pairs of factors to check.
So it is important that you attempt to factor out any
common factors first.
6x2y2 – 2xy2 – 60y2 = 2y2(3x2 – x – 30)
The only possible factors for 3 are 1 and 3, so we know
that, if we can factor the polynomial further, it will have to
look like 2y2(3x
)(x
) in factored form.
Continued.
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Factoring Polynomials
Example Continued
Since the product of the last two terms of the binomials
will have to be –30, we know that they must be
different signs.
Possible factors of –30 are {–1, 30}, {1, –30}, {–2, 15},
{2, –15}, {–3, 10}, {3, –10}, {–5, 6} or {5, –6}.
We will be looking for a combination that gives the sum
of the products of the outside terms and the inside terms
equal to –x.
2
Answer: 2y (3x - 10)(x + 3)
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Factoring Polynomials
Example Continued
Check the resulting factorization using the FOIL method.
F
O
I
L
(3x – 10)(x + 3) = 3x(x) + 3x(3) – 10(x) – 10(3)
= 3x2 + 9x – 10x – 30
= 3x2 – x – 30
So our final answer when asked to factor the polynomial
6x2y2 – 2xy2 – 60y2 will be 2y2(3x – 10)(x + 3).
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Home-Learning Assignment #3:
Page 508-510
(1-5, 8, 14, 22, 28, 51)
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