SOLUTION: Graph the polynomial function P(x) = x^4 + x^3 -3x^2- 5x -2 to approximately find the function's zeros, then use synthetic division and the remainder theorem to exactly find its ze

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Graph the polynomial function P(x) = x^4 + x^3 -3x^2- 5x -2 to approximately find the function's zeros, then use synthetic division and the remainder theorem to exactly find its ze      Log On


   

Question 90246: Graph the polynomial function P(x) = x^4 + x^3 -3x^2- 5x -2 to approximately find the function's zeros, then use synthetic division and the remainder theorem to exactly find its zeros.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
When we graph, we find a zero at x=-1

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 the numerator to the right of the test zero.
-1|11-3-5-2
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1|11-3-5-2
|
1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 1)
-1|11-3-5-2
|-1
1

Add -1 and 1 to get 0. Place the sum right underneath -1.
-1|11-3-5-2
|-1
10

Multiply -1 by 0 and place the product (which is 0) right underneath the third coefficient (which is -3)
-1|11-3-5-2
|-10
10

Add 0 and -3 to get -3. Place the sum right underneath 0.
-1|11-3-5-2
|-10
10-3

Multiply -1 by -3 and place the product (which is 3) right underneath the fourth coefficient (which is -5)
-1|11-3-5-2
|-103
10-3

Add 3 and -5 to get -2. Place the sum right underneath 3.
-1|11-3-5-2
|-103
10-3-2

Multiply -1 by -2 and place the product (which is 2) right underneath the fifth coefficient (which is -2)
-1|11-3-5-2
|-1032
10-3-2

Add 2 and -2 to get 0. Place the sum right underneath 2.
-1|11-3-5-2
|-1032
10-3-20

Since the last column adds to zero, we have a remainder of zero. This means x%2B1 is a factor of x%5E4+%2B+x%5E3+-+3x%5E2+-+5x+-+2

Now lets look at the bottom row of coefficients:

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

x%5E3+-+3x+-+2


So %28x%5E4+%2B+x%5E3+-+3x%5E2+-+5x+-+2%29%2F%28x%2B1%29=x%5E3+-+3x+-+2


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Now use the same test zero (which is -1) and perform synthetic division on x%5E3+-+3x+-+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 1x%5E3 to -3x%5E1 there is a zero coefficient for x%5E2. This is simply because x%5E3+-+3x+-+2 really looks like 1x%5E3%2B0x%5E2%2B-3x%5E1%2B-2x%5E0
-1|10-3-2
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-1|10-3-2
|
1

Multiply -1 by 1 and place the product (which is -1) right underneath the second coefficient (which is 0)
-1|10-3-2
|-1
1

Add -1 and 0 to get -1. Place the sum right underneath -1.
-1|10-3-2
|-1
1-1

Multiply -1 by -1 and place the product (which is 1) right underneath the third coefficient (which is -3)
-1|10-3-2
|-11
1-1

Add 1 and -3 to get -2. Place the sum right underneath 1.
-1|10-3-2
|-11
1-1-2

Multiply -1 by -2 and place the product (which is 2) right underneath the fourth coefficient (which is -2)
-1|10-3-2
|-112
1-1-2

Add 2 and -2 to get 0. Place the sum right underneath 2.
-1|10-3-2
|-112
1-1-20

Since the last column adds to zero, we have a remainder of zero. This means x%2B1 is a factor of x%5E3+-+3x+-+2

Now lets look at the bottom row of coefficients:

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

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


So %28x%5E3+-+3x+-+2%29%2F%28x%2B1%29=x%5E2+-+x+-+2

Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve x%5E2-x-2=0 ( notice a=1, b=-1, and c=-2)

x+=+%28--1+%2B-+sqrt%28+%28-1%29%5E2-4%2A1%2A-2+%29%29%2F%282%2A1%29 Plug in a=1, b=-1, and c=-2



x+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%2A1%2A-2+%29%29%2F%282%2A1%29 Negate -1 to get 1



x+=+%281+%2B-+sqrt%28+1-4%2A1%2A-2+%29%29%2F%282%2A1%29 Square -1 to get 1 (note: remember when you square -1, you must square the negative as well. This is because %28-1%29%5E2=-1%2A-1=1.)



x+=+%281+%2B-+sqrt%28+1%2B8+%29%29%2F%282%2A1%29 Multiply -4%2A-2%2A1 to get 8



x+=+%281+%2B-+sqrt%28+9+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%281+%2B-+3%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%281+%2B-+3%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%281+%2B+3%29%2F2 or x+=+%281+-+3%29%2F2

Lets look at the first part:

x=%281+%2B+3%29%2F2

x=4%2F2 Add the terms in the numerator
x=2 Divide

So one answer is
x=2



Now lets look at the second part:

x=%281+-+3%29%2F2

x=-2%2F2 Subtract the terms in the numerator
x=-1 Divide

So another answer is
x=-1

So our solutions are:
x=2 or x=-1

===================================

Answer:

So our zeros are:
x=-1 (with a multiplicity of 3), x=2