-Polynomials

By: the unknowns-Ashley -erick-laura -pj

Table of Contents• Multiplying Polynomials• Dividing Polynomials• Factoring Polynomials (Ax2+Bx+C)

• When A=1• Using Grouping (When A=1)

• Factoring Difference of Squares

Multiplying Polynomials

• Multiplying with Monomials -take the first term and multiply it with the other terms in

the equation.Ex. 1 3x * 2 =6xEx. 2 3x(2x+4) =6x2+12x

• Multiplying with Bionomials -take the two terms and multiply it with the other terms in

the equation.Ex. 1 (x+1)(x+3) =x2+3x + x + 3 = x2+4x + 3 <----collect like terms.Ex. 2 (x+2)2

=(x+2)(x+2) =x2+2x + 2x + 4 =x2+4x+4

Remember FOIL.F irst termsO uter termsI nner termsL ast termsThis is the order you multiply the terms by.

Dividing PolynomialsDividing polynomials is much like dividing

normal numbers.Ex.1 4x

2 Divide the like terms. 4/2 =2Move in any unused terms. 2x

24x

= 2x

Ex.2

12x+84

Divide each term by the dividend (bottom number).

12x4 + 8

4

3x+212x+84

=

Ex.3x3+4x2+7x+4

x+11.Change this so that it is in the form of a long division question.

x+1|x3+4x2+7x+4

Click the button to see the next steps.

Ex.3 cont.x+1|x3+4x2+7x+4

2.Take a look at the “x” term in the divisor. What do you need to multiply the “x” term by to get the first term in the dividend(x3)?

xm*xn=xm+n

x* =x3

3.Insert the answer into the question like so: x+1 x3+4x2+7x+4

x2

4. Multiply your divisor by the answer you got and add in the answer to the question.

x3

5. Subtract.

+ x2

3x2

6. Bring down the next term.

+7x

7. Look back at the “x” term. Find what you multiply the “x” term by to get the answer obtained in step 5. x*3x=3x2

8. Repeat steps 3-6.

+3x

3x2+3x4x +4

9. Repeat steps 7 and 8 until you cannot bring down anymore terms.

+4

+44x0

10. You now have your answer. If there is a remainder, put it at the end as R ___.

Tips

• When you re-write the question and change it into long division form, make sure the terms are in descending powers of “x”. Ex. x4+x3+x2+x+1

• If there is a “missing term” (Ex. x4+x3+ x+1 x2 is “missing”), add it in as 0xm remembering to put it in descending powers of “x”.

• Don’t forget that when you subtract a negative integer, you have to “flip” the sign and add. Ex. 2-(-2)=2+2=4

Factoring 1:(ax2+bx+c)Factors - Numbers or variables that multiply to get another value.

Ex #1: 12The factors of 12 are: 1x12, 3x4, 2x6, -1x-12, -3x-4, -2x-6Ex #2: x²+3x+2 “a” value=1 “b” value=3 “c” value =21. Find 2 numbers that multiply to give the “c” value and add up to give the “b” value. In this case, the only factors of the “c” value, 2, are 2 and 1. 2 and 1 add up to get 3, the “b” value.

2. Since the only factors of “x2” are “x” and “x”, we can put that into our answer:

(x )(x )

3. Now we can input the numbers we got from step 2.

(x+2)(x+1)

4. Since we can’t factor it any further, this becomes our answer.

x²+3x+2=(x+2)(x+1)

Notes:

-Remember that you can use negative numbers to find the 2 numbers for step 2.

-If you end up with an answer such as (x+2)(x+2), you can simplify it as (x+2)2.

Factoring 2: ( ax² + bx + c ) Ex. 2x² + 5x +3

If the “a” value doesn’t equal 1, then here’s what you do:

1. Multiply the “a” value and the “c” value.

2. Find the two numbers that multiply together to get “ac” and add together to get “b”. In this case, the numbers would be 2 and 3.

3. Rewrite the question, so that the “bx” part is written as “b1x+b2x”.

4. Factor the first two terms.

Then the last two terms.

5. The terms in the parentheses should be the same.

6. The part in the parentheses gives you half of your answer.

The other half comes from the terms outside of the parentheses.

You now have your answer.

6

2 3+ =5

2x² + 5x +3=2x2+2x+3x+3=2x(x+1)+3(x+1)

=(2x+3)(x+1)

Note: If one (or both) of the parentheses can still be factored, factor it. Ex: (4x+8)(2x+1) = 4(x+2)(2x+1)

Factoring 3: Difference of Squares• This is the easiest type of factoring.

25x2-49

All questions that involve this type of factoring come in this form, or a variation of this form.

First, find the square root of the first term. In this case, it would be “5x”. This would be the first terms in both of your parentheses.

(5x )(5x )

Next, find the square root of the second term. In this case, it would be 7. Since the number in the question is a negative number, when you insert the seven into the parentheses, put one as “+7” and the other as “-7”.

(5x+7)(5x-7)Now you have your answer.

25x2-49=(5x+7)(5x-7)

Note:

Don’t get confused if the question is something like 36x6-16y4. Just find the square root of the coefficient and the exponent of the first term to get the first half of your parentheses and the square root of the coefficient and the exponent of the second term to get the second half of your parentheses. In this case the answer would be (6x3+4y2)(6x3-4y2).

Also, don’t forget to make one set have a positive second term and the other set have a negative second term.