# 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.

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> 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.      Log On

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Question 193068: 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.

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# 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)