SOLUTION: I need help with this question: use synthetic division to find the polynomial which is not a factor of x^3+2x^2-9x-18. (x+2)(x-2)(x-3)(x+3)

Question 86300: I need help with this question: use synthetic division to find the polynomial which is not a factor of x^3+2x^2-9x-18. (x+2)(x-2)(x-3)(x+3)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Lets test the factor

Start with the given polynomial

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.

-2	\|	1	2	-9	-18
	\|

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

-2	\|	1	2	-9	-18
	\|
		1

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

-2	\|	1	2	-9	-18
	\|		-2
		1

Add -2 and 2 to get 0. Place the sum right underneath -2.

-2	\|	1	2	-9	-18
	\|		-2
		1	0

Multiply -2 by 0 and place the product (which is 0) right underneath the third coefficient (which is -9)

-2	\|	1	2	-9	-18
	\|		-2	0
		1	0

Add 0 and -9 to get -9. Place the sum right underneath 0.

-2	\|	1	2	-9	-18
	\|		-2	0
		1	0	-9

Multiply -2 by -9 and place the product (which is 18) right underneath the fourth coefficient (which is -18)

-2	\|	1	2	-9	-18
	\|		-2	0	18
		1	0	-9

Add 18 and -18 to get 0. Place the sum right underneath 18.

-2	\|	1	2	-9	-18
	\|		-2	0	18
		1	0	-9	0

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 3 coefficients (1,0,-9) form the quotient

So

----------------------------------------------------------------------
Lets test the factor

Start with the given polynomial

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.

2	\|	1	2	-9	-18
	\|

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

2	\|	1	2	-9	-18
	\|
		1

Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is 2)

2	\|	1	2	-9	-18
	\|		2
		1

Add 2 and 2 to get 4. Place the sum right underneath 2.

2	\|	1	2	-9	-18
	\|		2
		1	4

Multiply 2 by 4 and place the product (which is 8) right underneath the third coefficient (which is -9)

2	\|	1	2	-9	-18
	\|		2	8
		1	4

Add 8 and -9 to get -1. Place the sum right underneath 8.

2	\|	1	2	-9	-18
	\|		2	8
		1	4	-1

Multiply 2 by -1 and place the product (which is -2) right underneath the fourth coefficient (which is -18)

2	\|	1	2	-9	-18
	\|		2	8	-2
		1	4	-1

Add -2 and -18 to get -20. Place the sum right underneath -2.

2	\|	1	2	-9	-18
	\|		2	8	-2
		1	4	-1	-20

Since the last column adds to -20, we have a remainder of -20. This means is a not factor of
Now lets look at the bottom row of coefficients:

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

and the last coefficient -20, is the remainder, which is placed over like this

Putting this altogether, we get:

So

which looks like this in remainder form:
remainder -20

So the expression is not a factor of . This means the answer is
----------------------------------------------------------------------
Lets test the factor

Start with the given polynomial

First lets find our test zero:

Set the denominator equal to zero
Solve for x.

so our test zero is 3

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.

3	\|	1	2	-9	-18
	\|

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

3	\|	1	2	-9	-18
	\|
		1

Multiply 3 by 1 and place the product (which is 3) right underneath the second coefficient (which is 2)

3	\|	1	2	-9	-18
	\|		3
		1

Add 3 and 2 to get 5. Place the sum right underneath 3.

3	\|	1	2	-9	-18
	\|		3
		1	5

Multiply 3 by 5 and place the product (which is 15) right underneath the third coefficient (which is -9)

3	\|	1	2	-9	-18
	\|		3	15
		1	5

Add 15 and -9 to get 6. Place the sum right underneath 15.

3	\|	1	2	-9	-18
	\|		3	15
		1	5	6

Multiply 3 by 6 and place the product (which is 18) right underneath the fourth coefficient (which is -18)

3	\|	1	2	-9	-18
	\|		3	15	18
		1	5	6

Add 18 and -18 to get 0. Place the sum right underneath 18.

3	\|	1	2	-9	-18
	\|		3	15	18
		1	5	6	0

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 3 coefficients (1,5,6) form the quotient

So

-----------------------------------------------------------------
Lets test the factor

Start with the given polynomial

First lets find our test zero:

Set the denominator equal to zero
Solve for x.

so our test zero is -3

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.

-3	\|	1	2	-9	-18
	\|

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

-3	\|	1	2	-9	-18
	\|
		1

Multiply -3 by 1 and place the product (which is -3) right underneath the second coefficient (which is 2)

-3	\|	1	2	-9	-18
	\|		-3
		1

Add -3 and 2 to get -1. Place the sum right underneath -3.

-3	\|	1	2	-9	-18
	\|		-3
		1	-1

Multiply -3 by -1 and place the product (which is 3) right underneath the third coefficient (which is -9)

-3	\|	1	2	-9	-18
	\|		-3	3
		1	-1

Add 3 and -9 to get -6. Place the sum right underneath 3.

-3	\|	1	2	-9	-18
	\|		-3	3
		1	-1	-6

Multiply -3 by -6 and place the product (which is 18) right underneath the fourth coefficient (which is -18)

-3	\|	1	2	-9	-18
	\|		-3	3	18
		1	-1	-6

Add 18 and -18 to get 0. Place the sum right underneath 18.

-3	\|	1	2	-9	-18
	\|		-3	3	18
		1	-1	-6	0

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 3 coefficients (1,-1,-6) form the quotient

So

--------------------------------------------------------------------------
Answer:
So to recap, the only expression that is not a factor is

Notice when we graph the function

we can clearly see that the polynomial is not a factor since is not a root
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