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 ->  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


   

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.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1

To factor 6x%5E3+%2B+31x%5E2+%2B+4x+-+5, we can use synthetic division


First, let's find our test zero:

x%2B5=0 Set the given factor x%2B5 equal to zero

x=-5 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 6x%5E3+%2B+31x%5E2+%2B+4x+-+5 to the right of the test zero.
-5|6314-5
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 6)
-5|6314-5
|
6

Multiply -5 by 6 and place the product (which is -30) right underneath the second coefficient (which is 31)
-5|6314-5
|-30
6

Add -30 and 31 to get 1. Place the sum right underneath -30.
-5|6314-5
|-30
61

Multiply -5 by 1 and place the product (which is -5) right underneath the third coefficient (which is 4)
-5|6314-5
|-30-5
61

Add -5 and 4 to get -1. Place the sum right underneath -5.
-5|6314-5
|-30-5
61-1

Multiply -5 by -1 and place the product (which is 5) right underneath the fourth coefficient (which is -5)
-5|6314-5
|-30-55
61-1

Add 5 and -5 to get 0. Place the sum right underneath 5.
-5|6314-5
|-30-55
61-10

Since the last column adds to zero, we have a remainder of zero. This means x%2B5 is a factor of 6x%5E3+%2B+31x%5E2+%2B+4x+-+5

Now lets look at the bottom row of coefficients:

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

6x%5E2+%2B+x+-+1


So 6x%5E3+%2B+31x%5E2+%2B+4x+-+5 factors to %28x%2B5%29%286x%5E2+%2B+x+-+1%29


In other words, 6x%5E3+%2B+31x%5E2+%2B+4x+-+5=%28x%2B5%29%286x%5E2+%2B+x+-+1%29


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





# 2


First lets find our test zero:

x%2B1=0 Set the denominator x%2B1 equal to zero

x=-1 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 x%5E4+-2x3+%2B+x%5E2+-4 to the right of the test zero.(note: remember if a polynomial goes from 1x%5E2 to -4x%5E0 there is a zero coefficient for x%5E1. This is simply because x%5E4+-+2x%5E3+%2B+x%5E2+-+4 really looks like 1x%5E4%2B-2x%5E3%2B1x%5E2%2B0x%5E1%2B-4x%5E0
-1|1-210-4
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1|1-210-4
|
1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is -2)
-1|1-210-4
|-1
1

Add -1 and -2 to get -3. Place the sum right underneath -1.
-1|1-210-4
|-1
1-3

Multiply -1 by -3 and place the product (which is 3) right underneath the third coefficient (which is 1)
-1|1-210-4
|-13
1-3

Add 3 and 1 to get 4. Place the sum right underneath 3.
-1|1-210-4
|-13
1-34

Multiply -1 by 4 and place the product (which is -4) right underneath the fourth coefficient (which is 0)
-1|1-210-4
|-13-4
1-34

Add -4 and 0 to get -4. Place the sum right underneath -4.
-1|1-210-4
|-13-4
1-34-4

Multiply -1 by -4 and place the product (which is 4) right underneath the fifth coefficient (which is -4)
-1|1-210-4
|-13-44
1-34-4

Add 4 and -4 to get 0. Place the sum right underneath 4.
-1|1-210-4
|-13-44
1-34-40

Since the last column adds to zero, we have a remainder of zero. This means x%2B1 is a factor of x%5E4+-+2x%5E3+%2B+x%5E2+-+4

Now lets look at the bottom row of coefficients:

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

x%5E3+-+3x%5E2+%2B+4x+-+4


So x%5E4+-+2x%5E3+%2B+x%5E2+-+4 factors to %28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29


In other words, x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29


Now let's use the factor x-2 to factor x%5E3+-+3x%5E2+%2B+4x+-+4


First lets find our test zero:

x-2=0 Set the denominator x-2 equal to zero

x=2 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 x%5E3+-+3x%5E2+%2B+4x+-+4 to the right of the test zero.
2|1-34-4
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
2|1-34-4
|
1

Multiply 2 by 1 and place the product (which is 2) right underneath the second coefficient (which is -3)
2|1-34-4
|2
1

Add 2 and -3 to get -1. Place the sum right underneath 2.
2|1-34-4
|2
1-1

Multiply 2 by -1 and place the product (which is -2) right underneath the third coefficient (which is 4)
2|1-34-4
|2-2
1-1

Add -2 and 4 to get 2. Place the sum right underneath -2.
2|1-34-4
|2-2
1-12

Multiply 2 by 2 and place the product (which is 4) right underneath the fourth coefficient (which is -4)
2|1-34-4
|2-24
1-12

Add 4 and -4 to get 0. Place the sum right underneath 4.
2|1-34-4
|2-24
1-120

Since the last column adds to zero, we have a remainder of zero. This means x-2 is a factor of x%5E3+-+3x%5E2+%2B+4x+-+4

Now lets look at the bottom row of coefficients:

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

x%5E2+-+x+%2B+2


So %28x%5E3+-+3x%5E2+%2B+4x+-+4%29%2F%28x-2%29=x%5E2+-+x+%2B+2


Basically x%5E3+-+3x%5E2+%2B+4x+-+4 factors to %28x-2%29%28x%5E2+-+x+%2B+2%29


So x%5E3+-+3x%5E2+%2B+4x+-+4=%28x-2%29%28x%5E2+-+x+%2B+2%29


This means that x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x%5E3+-+3x%5E2+%2B+4x+-+4%29 then becomes


x%5E4+-+2x%5E3+%2B+x%5E2+-+4=%28x%2B1%29%28x-2%29%28x%5E2+-+x+%2B+2%29


So all you have to do now is factor x%5E2+-+x+%2B+2 (I'll let you do that)