Transcript Factoring Polynomials - City Colleges of Chicago

MODULE 14
Review of Factoring
Factoring Polynomials
This process is basically the REVERSE
of the distributive property.
distributive property
( x  2)(x  5) 
x  3x 10
factoring
2
Factoring Polynomials
In factoring you start with a polynomial
(2 or more terms) and you want to rewrite it
as a product (or as a single term)
Three terms
x  3x 10  ( x  2)(x  5)
2
One term
Techniques of Factoring
Polynomials
1. Factoring Out the Greatest
Common Factor (GCF).
The GCF for a polynomial is the largest
monomial that divides each term of the
polynomial.
Factor out the GCF:
4y  2y
3
2
Factoring Polynomials - GCF
4y  2y
3
2
Write the two terms in the
form of prime factors…
22y y y 2 y y
2yy ( 2 y
They have in common 2yy
1)
 2 y (2 y 1)
2
This process is basically the reverse of the distributive property.
Check the work….
2 y (2 y  1)  4y  2y
2
3
2
Factoring Polynomials - GCF
3 terms
Factor the GCF:
4ab  12a b c  8ab c 
3
2
3 2
2
4 a b ( b - 3a c
One term
4 2
+
2
2b c
2
)
Factoring Polynomials - GCF
EXAMPLE:
5x(2 x  4)  3(2 x  4) 
(2 x  4) ( 5x - 3 )
Examples
Factor the following polynomial.
12 x  20 x  3  4  x  x  4  5  x  x  x  x
2
4
 4  x  x (3  5  x  x )
 4 x (3  5 x )
2
2
Examples
Factor the following polynomial.
15 x y  3 x y  3  5  x  y  3  x  y
3
5
2
4
3
5
 3  x  y (5  x  y  1)
2
4
 3 x 2 y 4 (5 xy  1)
2
4
Techniques of Factoring
Polynomials
2. Factoring a Polynomial with four or more
Terms by Grouping
x  3x  2 x  6 
There is no GCF for all
four terms.
x ( x  3)  2 ( x  3) 
In this problem we factor GCF
by grouping the first two
terms and the last two terms.
3
2
2
( x  3) ( x  2)
2
Example 1
Factor 4m + 8n +mp + 2np
= (4m + 8n) + (mp + 2np)
=
4 (m + 2n) + p (m + 2n)
=
(m + 2n) (4 + p)
Example 2
Factor ax – 5ay + 2x – 10y
= (ax – 5ay) + (2x – 10y)
=
a (x – 5y) + 2 (x – 5y)
=
(x – 5y) (a + 2)
Example 3
Factor ab + 3a – 3b – 9
= (ab + 3a) + (– 3b – 9)
=
a (b + 3) + (– 3) (b + 3)
=
(b + 3) (a – 3)
Techniques of Factoring
Polynomials
3. Factoring Trinomials.
x  5x  6
2
We need to find factors of 6
….that add up to 5
Since 6 can be written as the product of 2 and 3
and 2 + 3 = 5, we can use the numbers 2 and 3
to factor the trinomial.
Factoring Trinomials, continued...
x  5x  6
2
2x3=6
2+3=5
Use the numbers 2 and 3 to factor the trinomial…
) x

Write the parenthesis, with
An “x” in front of each.
(x
Write in the two numbers
we found above.
( x  2 ) x  3 
Factoring Trinomials, continued...
So we factored the trinomial…
x  5 x  6  ( x  2 ) x  3 
2
You can check your work by multiplying back
to get the original answer
( x  2 ) x  3   x  3 x  2 x  6 
2
 x  5x  6
2
Factoring Trinomials
x  7x  6
2
Find factors of 6 that add up to 7
6 and 1
x  5x  6
2
Find factors of – 6 that add up to –5
– 6 and 1
x  1x  6
2
Find factors of – 6 that add up to 1
3 and –2
Factoring Trinomials
x  7x  6
2
( x  6 ) x  1 
factors of 6 that add up to 7:
x  5x  6
2
6
x  1x  6
1
( x  6 ) x  1 
factors of – 6 that add up to – 5: – 6
2
and
and
1
( x  3 ) x  2 
factors of – 6 that add up to 1:
3
and – 2
Factoring Trinomials
The hard case – “Box Method”
2x  x  6
2
Note: The coefficient of x2 is different from 1. In this case it is 2
2 x  1x  6
2
First: Multiply 2 and –6:
2 (– 6) = – 12
Next: Find factors of – 12 that add up to 1
– 3 and 4
Factoring Trinomials
The hard case – “Box Method”
2x  x  6
2
1. Draw a 2 by 2 grid.
2. Write the first term in the upper left-hand corner
3. Write the last term in the lower right-hand corner.
2x
2
6
Factoring Trinomials
The hard case – “Box Method”
2x  x  6
2
Find factors of – 12 that add up to 1
– 3 x 4 = – 12
–3+4=1
1. Take the two numbers –3 and 4, and put them, complete
with signs and variables, in the diagonal corners, like this:
2
2x
4x
–3 x
6
It does not matter which
way you do the diagonal
entries!
The hard case – “Box Method”
1. Then factor like this:
Factor Top Row
x 2x
2
4x
 3x
6
From Left Column
2x
2
x 2x
2 4x
 3x
6
Factor Bottom Row
x 2x
2
2
4x
 3x
6
From Right Column
2x
2
x 2x
2 4x
3
 3x
6
The hard case – “Box Method”
3
 3x
6
2x
2
x 2x
+2 +4 x
Note: The signs for the bottom row
entry and the right column entry
come from the closest term that you
are factoring from.
DO NOT FORGET THE SIGNS!!
Now that we have factored our box we can read off
our answer:
2 x  x  6  ( x  2)(2 x  3)
2
The hard case – “Box Method”
4 x  19 x  12 
2
Look for factors of 48 that add up to –19
x
2
4 x 4x
3
 3x
– 16 and – 3
4
 16 x
12
4 x 19x  12  (4 x  3)( x  4)
2
Finally, you can check your work by multiplying
back to get the original answer.
Use “Box” method to factor the
following trinomials.
1.
2x2 + 7x + 3
2.
4x2 – 8x – 21
3.
2x2 – x – 6
Check your answers.
1. 2x2 + 7x + 3 = (2x + 1)(x + 3)
2. 2x2 – x – 6 = (2x + 3)(x – 2)
3. 4x2 – 8x – 21 = (2x – 7)(2x + 3)
Note…
Not every quadratic expression can be
factored into two factors.
• For example x2 – 7x + 13.
We may easily see that there are no factors
of 13 that added up give us –7
• x2 – 7x + 13 is a prime trinomial.
Factoring the Difference of Two
Squares
(a + b)(a – b) = a2– ab + ab – b2 = a2 – b2
FORMULA:
a2 – b2 = (a + b)(a – b)
The difference of the squares of two quantities always
factors into the sum of these two quantities multiplied
by the difference of these two quantities.
Factoring the difference of two squares
2
a
Factor
Difference
of two squares
–
= (a + b)(a – b)
2
b
x2 – 4y2
(x)
2
2
(2y)
(x – 2y)(x + 2y)
Now you can check the results…
Factor 16r2 – 25
Difference
Of two squares
2
(4r)
2
(5)
(4r – 5)(4r + 5)
More Examples
y – 16 = (y)
2

2
2
(4) = (y – 4)(y + 4)
2
2
2
25x – 81 = (5x) – (9)
= (5x – 9)(5x +9)
4
2
2 2
2
x – 100y = (x ) – (10y)
2
2
= (x – 10y)(x +10y)
Factoring the Sum Two Cubes
Using the following formula:
a  b  (a  b)(a  ab  b )
3
3
2
2
The following polynomial can be
factored:
n 27
3
=
(n + 3) (n 3n  9)
2
Example 1
8n 125 = (2n)  (5)
2
= (2n + 5) [(2n) (2n)(5)  25]
2
= (2n + 5) (4n 10n  25)
3
3
3
Example 2
Factor
=
=
216r  125t
3
3
(6r )  (5t )
3
3
(6r  5t )[(6r )  (6r )(5t )  (5t ) ]
2
2
= (6r  5t )(36r 2  30rt  25t 2 )
Factoring the Difference of Two Cubes
Using the following formula:
a  b  (a  b)(a  ab  b )
3
3
2
2
The following polynomial can be factored:
n 27 = (n – 3) (n 3n  9)
3
2
Example 1
Factor 27m 1 = (3m)  (1)
3
3
=
3
(3m  1)(9m  3m  1)
2
Example 2
Factor x  1000y
3
3
= ( x)  (10y)
3
3
= ( x  10y)(x  10xy  100y )
2
2