SOLUTION: THE FACTOR THEOREM Factor P(x) = 6x^3 + 31x^2 + 4x -5 given that x+5 is one factor. Factor R(x) = x^4 -2x3 + x^2 -4, given that x+1 and x-2 are factors.

You can put this solution on YOUR website!
# 1

To factor , we can use synthetic division


First, let's find our test zero:

Set the given factor equal to zero

Solve for x.

so our test zero is -5


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

-5	|	6	31	4	-5
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)

-5	|	6	31	4	-5
	|
		6

Multiply -5 by 6 and place the product (which is -30) right underneath the second coefficient (which is 31)

-5	|	6	31	4	-5
	|		-30
		6

Add -30 and 31 to get 1. Place the sum right underneath -30.

-5	|	6	31	4	-5
	|		-30
		6	1

Multiply -5 by 1 and place the product (which is -5) right underneath the third coefficient (which is 4)

-5	|	6	31	4	-5
	|		-30	-5
		6	1

Add -5 and 4 to get -1. Place the sum right underneath -5.

-5	|	6	31	4	-5
	|		-30	-5
		6	1	-1

Multiply -5 by -1 and place the product (which is 5) right underneath the fourth coefficient (which is -5)

-5	|	6	31	4	-5
	|		-30	-5	5
		6	1	-1

Add 5 and -5 to get 0. Place the sum right underneath 5.

-5	|	6	31	4	-5
	|		-30	-5	5
		6	1	-1	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 (6,1,-1) form the quotient




So factors to


In other words,


I'll let you continue the factorization....

---

# 2


First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -1


Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of 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

-1	|	1	-2	1	0	-4
	|

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

-1	|	1	-2	1	0	-4
	|
		1

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

-1	|	1	-2	1	0	-4
	|		-1
		1

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

-1	|	1	-2	1	0	-4
	|		-1
		1	-3

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

-1	|	1	-2	1	0	-4
	|		-1	3
		1	-3

Add 3 and 1 to get 4. Place the sum right underneath 3.

-1	|	1	-2	1	0	-4
	|		-1	3
		1	-3	4

Multiply -1 by 4 and place the product (which is -4) right underneath the fourth coefficient (which is 0)

-1	|	1	-2	1	0	-4
	|		-1	3	-4
		1	-3	4

Add -4 and 0 to get -4. Place the sum right underneath -4.

-1	|	1	-2	1	0	-4
	|		-1	3	-4
		1	-3	4	-4

Multiply -1 by -4 and place the product (which is 4) right underneath the fifth coefficient (which is -4)

-1	|	1	-2	1	0	-4
	|		-1	3	-4	4
		1	-3	4	-4

Add 4 and -4 to get 0. Place the sum right underneath 4.

-1	|	1	-2	1	0	-4
	|		-1	3	-4	4
		1	-3	4	-4	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 4 coefficients (1,-3,4,-4) form the quotient




So factors to


In other words,


Now let's use the factor to factor


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 to the right of the test zero.

2	|	1	-3	4	-4
	|

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

2	|	1	-3	4	-4
	|
		1

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

2	|	1	-3	4	-4
	|		2
		1

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

2	|	1	-3	4	-4
	|		2
		1	-1

Multiply 2 by -1 and place the product (which is -2) right underneath the third coefficient (which is 4)

2	|	1	-3	4	-4
	|		2	-2
		1	-1

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

2	|	1	-3	4	-4
	|		2	-2
		1	-1	2

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

2	|	1	-3	4	-4
	|		2	-2	4
		1	-1	2

Add 4 and -4 to get 0. Place the sum right underneath 4.

2	|	1	-3	4	-4
	|		2	-2	4
		1	-1	2	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,-1,2) form the quotient




So


Basically factors to


So


This means that then becomes





So all you have to do now is factor (I'll let you do that)