Transcript Factoring by using different methods

Factoring by using different
methods
Factoring a quadratic without
2
a number in front of “x ”
The guess method
Ex: x2+4x-5
1. Find two numbers that multiply to give
you the last number and also add to
give you the middle number. The
second number is +4 and the last
number is -5.
Ex: x2+4x-5
2. The two numbers would be
“+5” and “-1”.
Since 5•-1= -5 and 5+(-1)=4.
3. Make the two sets of parenthesis and
insert the two numbers you found.
When there is no number in front of “x2”,
the parenthesis start out with “x”
(x+5)(x-1)
Using the quadratic formula
Ex: x2+4x-5
1. Label “a”,”b”, and “c”
a=1, b=4, c=-5
2. Plug in the numbers and solve.
(Note, this is only useful if you are finding
x-intercepts or solving the equation.)
Factor by grouping
• Used when there are four terms in the
polynomial.
• Goal: create two binomials out of the
polynomial by using other factoring
methods.
• Check: FOIL the two binomials and if
the product is the beginning polynomial,
the two binomials are correct.
Factor by Grouping
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Example of a polynomial:
(10x2 + 5x – 2x – 1)
Change to (10x2 +5x)(-2x -1)
Take GCF of both binomials.
5x(2x + 1)-1(2x+1) (Notice that the
remaining binomials are the same)
• Keep the identical binomial
• (5x-1)(2x+1)
Difference of squares
• Most will look like: (a2 – b2)
• Factored form: (a – b)(a + b)
• Hints: Be able to identify the squared
forms of numbers, such as :22 = 4, 32 =
9, 42=16… etc.
Other special Factoring
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Cubic Factors: (a3 ± b3)
Two factored forms:
(a3 + b3) = (a + b)(a2 – ab + b2)
(a3 – b3) = (a – b)(a2 + ab + b2)
• Helpful hint: know the cubed form of
numbers: 23 = 8, 33 = 27, 43 = 64, etc.
Factoring quadratic equations with a
number in front of the “x2” term.
1. By the guess method
2. By the a-c method
3. With the quadratic formula
Using the guess method
Ex. 4x2-12x+9
1. Find the factors of the first and last
terms.
2. Factors of the first term are (4x and x)
or (2x and 2x), factors of the last term
are (9 and 1) or (3 and 3)
3. Use those factors to build the two sets
of parenthesis.
4. Foil the answer you got to see if it is
correct. If it isn’t, choose a different
combination.
3. Use those factors to build the two sets
of parenthesis. For example,
(2x+3)(2x+3)
4.Foil the answer you got to see if it is
correct. If it isn’t, choose a different
combination.
Using the a-c method
Ex. 6x2-5x-6
1. Multiply “a” by “c”.
a = 6 b = -5 c = -6
Using the a-c method
Ex. 6x2-5x-6
1. Multiply “a” by “c”.
a = 6 b = -5 c = -6
6•-6 = 36
2. Find two numbers that multiply to “a•c”
and add up to give you “b”. An easy
way to do this is to list the factors of
“a•c”.
a•c = -36
b = -5
What are the numbers?
2. Find two numbers that multiply to “a•c”
and add up to give you “b”. An easy
way to do this is to list the factors of
“a•c”.
a•c = -36
b = -5
What are the numbers?
(-9)•(4) = -36
(-9)+(4) = -5
3. Take the two numbers that we found in
part 2 and substitute them in for the
“-5x” in the original equation, resulting
in
6x2-9x+4x-6
6x2-9x+4x-6
4. Since there are now four terms, we will
use factoring by grouping. So, break
the equation apart.
(6x2-9x)(4x-6)
(6x2-9x)(4x-6)
5. Factor each set of parenthesis
separately. The terms inside both
parenthesis have to match.
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(6x2-9x)(4x-6)
Factor each set of parenthesis
separately. The terms inside both
parenthesis have to match.
3x(x-3) and +2(x-3)
3x(x-3) and +2(x-3)
6. Combine the two outside terms in a
parenthesis and keep one of the
matching sets in another.
(3x+2)(x-3)
Using the quadratic formula
Ex. 6x2-5x-6
1. Locate the three numbers for “a”,”b”,
and “c”.
a = 6 b = -5 c = -6
2. Plug the values into the quadratic
formula.
3. solve