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Factoring Polynomials

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Rob Frederick

on 21 February 2013

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Transcript of Factoring Polynomials

Factoring by GCF Factoring Polynomials Factoring ax + bx + c 2 Factoring x + bx + c 2 Factoring Special Products If f(x) = x, what best describes the relationship between x and f(x)?

A.) As the value of x increases by 1, the value of f(x) will increase by 2.

B.) As the value of x increases by 1, the value of f(x) will remain the same.

C.) As the value of x increases by 1, the value of f(x) will increase by .

D.) The value of x will not affect the value of f(x), Bell Ringer 2 1 1 2 _ _ Objectives 1.) Write the prime factorization of numbers.

2.) Find the GCF of monomials.

3.) Factor a polynomial by GCF PASS 1.1.d: Solve two-step and three-step problems using concepts such as rules of exponents, rate, distance, ratio and proportion, and percent. Definitions The numbers that are multiplied to find a product are called the factors of that product. A number is divisible by its factors.

The set of prime numbers that multiply to form a product is called the prime factorization of that product. Each number has only one prime factorization.

*Remember that prime numbers are numbers that are divisible only by one and themselves (2, 3, 5, 7, 11, 13, 17, 19, 23...)

Factors that are shared between two numbers are called common factors, and the greatest of these is the greatest common factor (also known as the GCF). Today we will talk about how to find a GCF. Prime Factorization Before we find the GCF, we have to be able to factor whole numbers completely, or write their prime factorization. There are two common ways to do this: a factor tree or a ladder diagram. Write the prime factorization of 120. 120 120= 2 2 2 5 3 or 2 3 5 Factor Tree 2 60 2 30 10 3 2 5 120 Ladder Diagram 2 60 2 30 10 3 2 5 1 5 Choose any two factors of the product. Then keep finding factors until each branch ends in a prime factor Factor Tree Choose a prime factor of the product and keep dividing by prime numbers until the quotient is one. Ladder Diagram . . . . . . 3 40 Write the prime factorization: 33 40 = 2 5 Write the prime factorization: 33 = 3 11 3 . . GCF Finding the GCF has a few steps.
1.) List the prime factorization.
2.) Align common factors
3.) Multiply the common factors to find the GCF.

*Sometimes step 2 will reveal only one common factor. In this case, it is the GCF.

*Occasionally the GCF of two numbers is 1. In this case the two numbers are relatively prime. 24 and 60 Find the GCF: factors of 24: 2 2 2 3 factors of 60: 2 2 3 5 The GCF of 24 and 60 is 12. 18 and 27 factors of 18: 2 3 3 . . . . . . . . 2 2 3 = 12 factors of 27: 3 3 3 . . . . 3 3 = 9 . The GCF of 18 and 27 is 9. 12 and 16 Find the GCF The GCF 12 and 16
is 4. GCF of Monomials The GCF of monomials is found the same way as the GCF of numbers, except that you have to account for any shared variables. Find the GCF of 3x and 6x . 3 2 3x = 3 x x x
6x = 2 3 x x

3 x x = 3x

The GCF of 3x and 6x is 3x . 1.) Write the prime factorization 2.) Align common factors 3.) Multiply the common factors 2 2 2 3 3 . . . . . . . . 2 18g and 27g h. Find the GCF. 2 3 The GCF of 18g and 27g h is 9g . Find the GCF. 2 2 3 Factoring by GCF Sometimes, when polynomials have terms with a GCF, we can factor or "take apart" the polynomial into its factors. This is called factoring. Factor 10y + 20y - 5y. 1.) Find the GCF The GCF of 10y , 20y , and -5y is 5y. 2.) Factor the GCF
out of each term. 10y + 20y - 5y

2y (5y) + 4y(5y) - 1(5y)

5y(2y + 4y - 1) 3.) Factor out the GCF.
(This is the reverse of the Distributive Property.) *4.) You can then use the
Distributive Property to
check your work... 3 2 2 2 3 2 2 3 8x + 4x - 2x Factor using GCF 2 3 4 2x (4x + 2x - 1) Factor using GCF 2 2 PASS 1.2.c: Factor polynomial expressions. Bell Ringer On Saturday, Suzie's Pretzel Stand sells a total of 12 items. Suzie charges \$3 for a pretzel and \$2 for a milkshake. If she took in \$32 on Saturday, which system of equations can be used to determine how many pretzels and how many milkshakes were sold?

A.) 3x + 2y = 12 B.) 3x - 2y = 32
x + y = 32 x + y = 12

C.) 3x - 2y = 12 D.) x + y = 12
x + y = 32 3x + 2y = 32 { { { { We just spent some time factoring polynomials
by GCF. Try a practice problems to make sure
you remember... Review of Factoring by GCF 14x + 63x - 7x 3 2 Solution 7x(2x + 9x - 1) 2 Objectives Factor quadratic trinomials of the form x + bx + c. 2 PASS 1.2.c: Factor polynomial expressions. We multiplied binomials using FOIL. Remember Chapter 7... (x + 2)(x + 5) = x + 7x + 10 2 Look for a pattern... Notice that when we multiplied the constants in the binomials, the product was the constant in the trinomial, and when we added constants, the sum was the middle term. This will always be true whenever x has a coefficient of 1. 2 Guess-and-Check Method Using what we've just talked about, let's try factoring x + 19x + 60 2 ( + )( + ) First, we need the two sets of parenthesis... And, now we should be able to figure out the first terms... Since x x = x ,
the first terms must be x. ( x + )( x + ) Now, we have to figure out
the last terms... Remember in our example, the last terms multiplied to be the constant term, and added to be the middle term. (x + 2)(x + 5) = x + 7x + 10 2 (x + )(x + ) = x + 19x + 60 2 What does that mean for our problem? The last terms are
up to 19, but multiply
together to be 60... So we need to try
factors of 60 until
we find two that
add up to 19. the only possible solution is... In the end, ( x + 4 )( x + 15) Since 4 and 15 are the only factors
of 60 that also have a sum of 19. . 2 The Box Method Remember we can also use the box method to multiply
polynomials. So we can use the same "box" to help us factor x + 6x + 8 2 First, we start with a 2 x 2 box, since we know this
is a product of 2 binomials... Then, we have to fill in the parts
of the trinomial we know for sure... The solution will be the binomials
on the top and side of the box. And from that, we now the
first terms of the binmials are "x," We can fill in the first
and last terms of the
trinomial... x x x 8 and the last
two are factors
of the product "8". 2 8 and 1 OR 2 and 4 This means the
last terms are either x x x 8 2 x x x 8 2 8 1 2 4 8x x 2x 4x How do we tell which is
the correct answer? (x + 8)(x + 1) (x + 2)(x + 4) + + + + Only one of our solutions checks out. Remember the
original problem... 8(1) = 8
8 + 1 = 9 2(4) = 8
2 + 4 = 6 x + 6x + 8 2 So our last terms had a
product of 8 and a sum of 6. Therefore, (x + 2)(x + 8)
is the factored version
of our polynomial. A Quick Note About Signs... The signs of the original trinomial* give us clues about what the signs in the binomial factors are:

1.) IF BOTH signs in the trinomial are POSITIVE,
then BOTH signs in the binomial factors are POSITIVE.

2.) IF the FIRST sign of the trinomial is NEGATIVE AND the SECOND sign is POSITIVE, then BOTH signs of the binomial factors are NEGATIVE. one sign is positive
and
one sign is negative. In all other cases... Practice 1.) x + 13x + 30
2.) x - 6x + 5
3.) x + 4x - 21
4.) x - 10x - 24 Factor each trinomial (either by guess-and-check or the box method). You may check your answer by FOIL/Box. 2 2 2 2 5.) x + 11x + 18
6.) x - 11x + 18
7.) x + 11x + 10
8.) x - 6x - 27 2 2 2 2 Remember you can ALWAYS
check your answers so you NEVER have to get one wrong. Practice Solutions 1.) (x + 3)(x + 10)
2.) (x - 2)(x -3)
3.) (x + 7)(x - 3)
4.) (x - 12)(x + 2) 5.) (x + 2)(x + 9)
6.) (x - 2)(x - 9)
7.) (x + 10)(x + 1)
8.) (x - 9)(x + 3) *We'll be discussing two methods:
1.) The Guess-and-check Method
2.) The Box Method 4(15) 1(60) 2(30) 3(20) 5(12) 6(10) *When the coefficient of x is 1. 2 "A plus in the back, two of what's in the front...
A minus in the back, one of each." Another easy way to
remember these rules... Bell Ringer What is the slope of the line shown?

A.) -4/3
B.) -3/4
C.) 3/4
D.) 4/3 Objectives Factor trinomials in the form

ax + bx + c 2 PASS 1.2.c: Factor polynomial expressions. We factored trinomials when the
leading coefficient was one... Previously.... x - 13x + 42 2 (x - 6)(x - 7) Remember that the constants in the
factors were factors of the constant
in the trinomial, and have a sum
that is the coefficient of
the middle term. We will look at problems like this: Today... Factor. 4x + 16x + 15 2 4x + 16x + 15 2 Guess-and-check Just like our last lesson, guess-and-check and the box method are valid options. But, since x-squared has a coefficient other than one, we have to account for the factors of its coefficient (in this case, 4), making it more difficult since there are more options to guess. We'll look at 3 methods:
1.) The Guess-and-check Method
2.) The Box Method
3.) The Illegal Move ( x + )( x + ) We could try
2x and 2x... OR 4x and x... We'll end up with the following: After guessing
and checking
with factors of
4 and 15... (4x + 1)(x + 15)
(4x + 3)(x + 5)
(4x + 5)(x + 3)
(4x + 15)(x + 1)
(2x + 15)(2x + 1)
(2x + 5)(2x + 3) And we'll need to check each of these
by FOILing. Notice that the "F" in FOIL gives us the first term in the trinomial...

While the "OI" gives us the middle term when we add like terms...

And the "L" gives us the last term. We can use this to check in our heads or on paper
until we arrive at the only possible factors... And the answer is... (2x + 5)(2x + 3) = 4x + 16x + 15 2 F L O I The Box Method The box method still works the same way, except that, just like the guess-and-check method, a LOT of guessing until you get it right is required: 4x + 16x + 15 2 4x 2 15 We fill in the missing
pieces until we get: Eventually... 4x 2 15 2x 2x 5 3 6x 10x + + is there an easier way? You have to ask... While guess-and-check and the box method
aren't so bad when a (the coefficient of x-squared) is prime, they get kind of tedious any other time... Luckily, we can use the Illegal Move The Illegal Move Let's take a look at the illegal move using the problem we have already factored: 4x + 16x + 15

x + 16 + 60

(x + 10)(x + 6)

(2x + 5)(2x + 3) 2 2 1.) Multiply the lead coefficient by the constant.
(THIS is the illegal move...we'll fix it later.) 2.) Then we factor the much easier trinomial. 3.) Now we "put back" the factors of the original lead coefficient* dividing them out of the constants and putting them on the "opposite" parenthesis x (in this case 2 and 2.) 4.) Finally, we check our answer with FOIL. *Step 3 sometimes means we only need to divide once (like on a prime number such as 3), and sometimes twice (like above). 2 2 Factor by any method. Practice 1.) 2x + 11x + 12
2.) 5x - 14x + 8
3.) 4y + 7y - 2 4.) 4x + 19x - 5
5.) 2x - 7x - 15
6.) -2x - 15x - 7 *NOTE: When the leading coefficient is negative, make sure to factor out the GCF of -1 first. In fact, it's ALWAYS best practice to factor ANY GCF you first. You should have gotten: Practice Solutions 1.) (x + 4)(2x + 3)
2.) (x - 2)(5x - 4)
3.) (y + 2)(4y - 1) 4.) (x + 5)(4x - 1)
5.) (x - 5)(2x + 3)
6.) -1(x + 7)(2x + 1) Don't forget to check your answers. FOIL! ! 2 2 2 2 2 2 (0, 2) (4, -1)
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