SOLUTION: Please help me out, I don't know how to begin to approach these. :[ Thanks so much Divide using synthetic division. a) (x^2 + 7x + 12) divided by (x + 4) b) (x^4 - 7x^2 + 9x

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Question 147620: Please help me out, I don't know how to begin to approach these. :[
Thanks so much
Divide using synthetic division.
a) (x^2 + 7x + 12) divided by (x + 4)
b) (x^4 - 7x^2 + 9x - 10) divided by (x - 2)
c) (2x^4 - 11x^3 + 15x^2 + 6x - 18) divided by (x - 3)
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you started.


a)



Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -4


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
-4|1712
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-4|1712
|
1

Multiply -4 by 1 and place the product (which is -4) right underneath the second coefficient (which is 7)
-4|1712
|-4
1

Add -4 and 7 to get 3. Place the sum right underneath -4.
-4|1712
|-4
13

Multiply -4 by 3 and place the product (which is -12) right underneath the third coefficient (which is 12)
-4|1712
|-4-12
13

Add -12 and 12 to get 0. Place the sum right underneath -12.
-4|1712
|-4-12
130

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 2 coefficients (1,3) form the quotient




So where

You can use this online polynomial division calculator to check your work





Let's simplify this expression using synthetic division


Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 2


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like
2|10-79-10
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
2|10-79-10
|
1

Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is 0)
2|10-79-10
|2
1

Add 2 and 0 to get 2. Place the sum right underneath 2.
2|10-79-10
|2
12

Multiply 2 by 2 and place the product (which is 4) right underneath the third coefficient (which is -7)
2|10-79-10
|24
12

Add 4 and -7 to get -3. Place the sum right underneath 4.
2|10-79-10
|24
12-3

Multiply 2 by -3 and place the product (which is -6) right underneath the fourth coefficient (which is 9)
2|10-79-10
|24-6
12-3

Add -6 and 9 to get 3. Place the sum right underneath -6.
2|10-79-10
|24-6
12-33

Multiply 2 by 3 and place the product (which is 6) right underneath the fifth coefficient (which is -10)
2|10-79-10
|24-66
12-33

Add 6 and -10 to get -4. Place the sum right underneath 6.
2|10-79-10
|24-66
12-33-4

Since the last column adds to -4, we have a remainder of -4. This means is not a factor of
Now lets look at the bottom row of coefficients:

The first 4 coefficients (1,2,-3,3) form the quotient



and the last coefficient -4, is the remainder, which is placed over like this





Putting this altogether, we get:



So where

which looks like this in remainder form:
remainder -4 where


You can use this online polynomial division calculator to check your work
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