1 5.1 Factoring a monomial from a polynomial Factors-numbers (or variables) that you multiply...
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Transcript of 1 5.1 Factoring a monomial from a polynomial Factors-numbers (or variables) that you multiply...
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5.1 Factoring a monomial from a polynomial
Factors-numbers (or variables) that you multiply together to get a product
The factors of 30 = 1,2,3,5,6,10,15,30
To factor something is the opposite of multiplying
Multiplying 3(x+2) = 3x + 6
Factoring 3x + 6 = 3(x+2)
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5.1 Factoring a monomial from a polynomial
GCF Greatest Common Factor
a number that will divide evenly into two or more numbers; the largest one you can find.
Find GCF of 48 and 60 is 12
Factors of 48=1,2,3,4,6,8,12,16,24,48
Factors of 60=1,2,3,4,5,6,10,12,15,20,30,60
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5.1 Factoring a monomial from a polynomialWe don’t want to have to list all the factors
every time we need to find a gcf. Sometimes we can tell by looking at the numbers. GCF of 12 and 16 is 4
If not use the prime factorization method:Make a factor tree for each number. Once
you have all primes, put your prime factorization into exponent form. Compare the two prime factorizations and see what they have in common. (Do 48 and 60 on board)
Now find the GCF of 18 and 24
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5.1 Factoring a monomial from a polynomial
You can apply this same idea to terms:
GCF of 18y2, 15y3, and 27y5
Is 3y2
GCF of 2(x+y) and 3x(x+y)
Is (x+y)
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5.1 Factoring a monomial from a polynomial
To factor a monomial from a polynomial,
1)Find GCF of all the terms
2) Put GCF out front
3) Use the dist property in reverse to factor out the GCF from EACH term
If you multiply it back in, you should get what you started with
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5.1 Factoring a monomial from a polynomial
15x – 20 = 5 (3x – 4)
6y2 + 9y5 = 3y2 (2 + 3y3)
35x2 – 25x + 5 = 5 (7x2 - 5x +1)
2x(x - 3) – 5(x - 3) = (x - 3) (2x - 5)
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5.3 Factoring Trinomials
This is getting ahead of us, but so you will understand where this fits in—
These are the steps we use to factor something completely
1) Check for a GCF (you learned in 5.1)
2) Count the terms
trinomial-FOIL backwards (5.3 and 5.4)
binomial-DOS or sum/diff of cubes
4 terms-grouping
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X2 + 7x + 12
GCF? No trinomial? Yes FOIL backwards:
X2 + 7x + 12 = ( )( )
FOIL start with F – to get X2 you need x,x
X2 + 7x + 12 = (x )(x )
FOIL now do L – to get 12, you could use any two factors of 12 (1,12 or 2,6 or 3,4)
which pair will add or subtract to 7?
1,12 = 13 or 11
2,6 = 8 or 4
3,4 = 7 or 1
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So 3,4 will work
X2 + 7x + 12 = (x 3)(x 4)
Now consider the signs:
If the constant (12) is positive, both factors will have the same sign-both (+) or both (-)
If the constant (12) is negative, the two factors will have different signs; one positive and one negative
In this case, in order to get a (+)12, we need both signs to be the same. In addition, we want a (+) 7 in the middle so we would choose both positives
X2 + 7x + 12 = (x + 3)(x + 4)
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X2 + 7x + 12 = (x + 3)(x + 4)
Lastly, check your smile lines
Assuming you chose wisely, you shouldn’t need to check x times x. it should equal x2;
You shouldn’t need to check 3 times 4. it should equal 12. But double check the middle term
X2 + 7x + 12 = (x + 3)(x + 4)
Can you see the 3x and 4x that will combine for a 7x in the middle?
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5.3 Factoring Trinomials
Factor X2 + x - 6 = (x + 3)(x - 2)
start with F
then L; consider 1,6 or 2,3
which will give you a 1 in the middle?
2,3 if we subtract them
since the constant is negative, we want our signs to be a plus and a minus
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5.3 Factoring Trinomials
Factor X2 + 2x - 24 = (x +6)(x - 4)
F: x and x
L: 1,24 or 2,12 or 3,8 or 4,6
Signs: -24 means you need a (+) (-)
Arrange them so the two will be (+)
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Trinomial signs binomial signs ax2 + bx + c ( + )( + )
ax2 - bx + c ( - )( - )
ax2 - bx - c ( + )( - ) and OR
ax2 + bx - c ( - )( + )
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5.3 Factoring TrinomialsFactor 2X2 + 2x - 12 Check for GCF first ALWAYS!!!GCF=2 so 2(X2 + x – 6)F: x and xL: 1,2 or 2,3Signs: -6 means you need a (+) (-)Arrange them so the one will be (+)2X2 + 2x - 12 2(X2 + x – 6)2(x + 3)(x - 2)
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5.4 Factoring Trinomials when a=1The same method applies here as in 5.3 but
you will notice that the coefficient on the x2
term will impact our middle term so finding the pair of factors becomes more challenging. You will often have to do more trial and error on paper to find the set that works. There are two approaches:
Guess and Check which follows immediately
Grouping which comes after guess/check
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2x2 + 13x + 15 = start with F, then L, then signs, check MT
(2x + 15)(1x + 1)15x + 2x =17x NO
(2x + 1)(1x + 15)1x + 30x = 31x NO
(2x + 5)(1x + 3)5x + 6x = 11x NO
(2x + 3)(1x + 5)3x + 10x = 13x YES!
Guess and Check
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2x2 + 7x + 6 = for 6, consider 1,6 and 2,3
(2x + 6)(1x + 1)
(2x + 1)(1x + 6)
(2x + 2)(1x + 3)
(2x + 3)(1x + 2)4x + 3x = 7x YES!
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3x2 + 20x + 12 = for 12, consider 1,12 and 2,6 and 3,4
(3x + 1)(1x + 12)
(3x + 12)(1x + 1)
(3x + 2)(1x + 6)
(3x + 6)(1x + 2)
(3x + 3)(1x + 4)
(3x + 4)(1x + 3)
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5x2 - 7x - 6 = for 6, consider 1,6 and 2,3
(5x 1)(1x 6)
(5x 6)(1x 1)
(5x 2)(1x 3)
(5x + 3)(1x - 2)
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8x2 + 33x + 4 = for 4, consider 1,4 and 2,2
for 8, consider 1,8 and 2,4
(8x + 4)(1x + 1)
(8x + 1)(1x + 4)
(8x + 2)(1x + 2)
(2x + 1)(4x + 4)
(2x + 4)(4x + 1)
(2x + 2)(4x + 2)
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Grouping2x2 + 13x + 15 where a=2 b=13 and c=15
Take a times c so 2(15) = 30
1) Find two numbers whose product is ac and whose sum is b
In this case, two numbers whose product is 30 and whose sum is 13
Consider the factors of 30:
1,30 and 2,15 and 3,10, and 5,6
3,10 add to 13
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2) Rewrite the middle term as the sum or difference of the two factors
2x2 + 13x + 15 2x2 + 10x + 3x + 15 3) Factor the new expression by
grouping two and two2x2 + 10x + 3x + 15 GCF of the 1st two? 2x of the 2nd two? 32x (x+5) + 3 (x+5)Pull the common factor to the front now(x+5) (2x+3) DONE!
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2x2 + 7x + 6
ac = 2(6) = 12 and b = 7
1,12 and 2,6 and 3,4
2x2 + 7x + 6
2x2 + 3x + 4x + 6
group
x(2x + 3) + 2(2x + 3)
Pull 2x+3 to the front
(2x + 3)(x + 2)
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3x2 + 20x + 12
ac = 3(12) = 36 and b = 20
1,36 and 2,18 and 3,12, and 4,9 and 6,63x2 + 20x + 12 3x2 + 18x + 2x + 12 group
3x(x + 6) + 2(x + 6)
Pull to the front
(x + 6)(3x + 2)
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5x2 - 7x - 6 ac = 5(-6) = -30 and b = -7
1,30 and 2,15 and 3,10 and 5,6
Consider the signs here 3 and -10 = -75x2 - 7x - 6 5x2 - 10x + 3x - 6group
5x(x - 2) + 3(x - 2)
Pull to the front
(x - 2)(5x + 3)
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8x2 + 33x + 4ac = 8(4) = 32 and b = 33
1,32 and 2,16 and 4,8 8x2 + 33x + 4 8x2 + 32x + 1x + 4 group
8x(x + 4) + 1(x + 4)
Pull to the front
(x + 4)(8x + 1)
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5.5 Special formulasRecall the factoring process
1) Check for a GCF (you learned in 5.1)
2) Count the terms
trinomial-FOIL backwards (5.3 and 5.4)
binomial-DOS or sum/diff of cubes(5.5)
4 terms-grouping
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Difference of squaresDOS always involves perfect square
numbers: like 1,4,9,16, 25, 36, 49, 64, etc.
It is always “something squared minus something squared”
a2 – b2 = (a + b)(a - b)
x2 – 16 = x2 – 42 = (x + 4)(x - 4)
25x2 – 4 = (5x)2 –(2)2 = (5x + 2)(5x - 2)36x2 – 49y2 = (6x)2 –(7y)2=(6x + 7y)(6x – 7y)
“sum of squares” are always prime.
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Difference of cubes/Sum of cubesDOC/SOC always involves perfect cube
numbers: like 1, 8, 27, 64, 125, 216, etc.
It is always “something cubed plus/minus something cubed”
a3 – b3 = (a - b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 - ab + b2)
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Difference of cubes
a3 – b3 = (a - b)(a2 + ab + b2)
y3 – 125 = y3 –53 = (y - 5)(y2 + y(5) + 52)
(y - 5)(y2 + 5y + 25)
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Sum of cubes
a3 + b3 = (a + b)(a2 - ab + b2)
8p3 + k3 = (2p)3 +(k)3 =
(2p + k)((2p)2 - (2p)(k) + k2)
(2p + k)(4p2 - 2pk + k2)