SOLUTION: Hi this is my problem: Find all real and imaginary roots: x^4+3x^3+x^2+4=0 Thank You!

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Question 110674This question is from textbook College Algebra
: Hi this is my problem:
Find all real and imaginary roots:
x^4+3x^3+x^2+4=0
Thank You! This question is from textbook College Algebra

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of 4 (the last coefficient):



Now let's list the factors of 1 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur





To save time, I'm only going to use synthetic division on the possible zeros that are actually zeros of the function.
Otherwise, I would have to use synthetic division on every possible root (there are 6 possible roots, so that means there would be at most 6 synthetic division tables).
However, you might be required to follow this procedure, so this is why I'm showing you how to set up a problem like this


If you're not required to follow this procedure, simply use a graphing calculator to find the roots


So it turns out that x=-2 is a root. So that means x=-2 is a test zero




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%5E2 to 4x%5E0 there is a zero coefficient for x%5E1. This is simply because x%5E4+%2B+3x%5E3+%2B+x%5E2+%2B+4 really looks like 1x%5E4%2B3x%5E3%2B1x%5E2%2B0x%5E1%2B4x%5E0
-2|13104
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-2|13104
|
1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 3)
-2|13104
|-2
1

Add -2 and 3 to get 1. Place the sum right underneath -2.
-2|13104
|-2
11

Multiply -2 by 1 and place the product (which is -2) right underneath the third coefficient (which is 1)
-2|13104
|-2-2
11

Add -2 and 1 to get -1. Place the sum right underneath -2.
-2|13104
|-2-2
11-1

Multiply -2 by -1 and place the product (which is 2) right underneath the fourth coefficient (which is 0)
-2|13104
|-2-22
11-1

Add 2 and 0 to get 2. Place the sum right underneath 2.
-2|13104
|-2-22
11-12

Multiply -2 by 2 and place the product (which is -4) right underneath the fifth coefficient (which is 4)
-2|13104
|-2-22-4
11-12

Add -4 and 4 to get 0. Place the sum right underneath -4.
-2|13104
|-2-22-4
11-120

Since the last column adds to zero, we have a remainder of zero. This means x%2B2 is a factor of x%5E4+%2B+3x%5E3+%2B+x%5E2+%2B+4

Now lets look at the bottom row of coefficients:

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

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


So %28x%5E4+%2B+3x%5E3+%2B+x%5E2+%2B+4%29%2F%28x%2B2%29=x%5E3+%2B+x%5E2+-+x+%2B+2

You can use this online polynomial division calculator to check your work

Basically x%5E4+%2B+3x%5E3+%2B+x%5E2+%2B+4 factors to %28x%2B2%29%28x%5E3+%2B+x%5E2+-+x+%2B+2%29

Now lets break x%5E3+%2B+x%5E2+-+x+%2B+2 down further

Again it turns out that x=-2 is a root of x%5E3+%2B+x%5E2+-+x+%2B+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|11-12
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-2|11-12
|
1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 1)
-2|11-12
|-2
1

Add -2 and 1 to get -1. Place the sum right underneath -2.
-2|11-12
|-2
1-1

Multiply -2 by -1 and place the product (which is 2) right underneath the third coefficient (which is -1)
-2|11-12
|-22
1-1

Add 2 and -1 to get 1. Place the sum right underneath 2.
-2|11-12
|-22
1-11

Multiply -2 by 1 and place the product (which is -2) right underneath the fourth coefficient (which is 2)
-2|11-12
|-22-2
1-11

Add -2 and 2 to get 0. Place the sum right underneath -2.
-2|11-12
|-22-2
1-110

Since the last column adds to zero, we have a remainder of zero. This means x%2B2 is a factor of x%5E3+%2B+x%5E2+-+x+%2B+2

Now lets look at the bottom row of coefficients:

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

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


So %28x%5E3+%2B+x%5E2+-+x+%2B+2%29%2F%28x%2B2%29=x%5E2+-+x+%2B+1

You can use this online polynomial division calculator to check your work

Basically x%5E3+%2B+x%5E2+-+x+%2B+2 factors to %28x%2B2%29%28x%5E2+-+x+%2B+1%29

Now lets break x%5E2+-+x+%2B+1 down further



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%2B1=0 ( notice a=1, b=-1, and c=1)




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



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



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



x+=+%281+%2B-+sqrt%28+-3+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%281+%2B-+i%2Asqrt%283%29%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-+i%2Asqrt%283%29%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=1%2F2+%2B+sqrt%283%29%2F2%2Ai or x=1%2F2+-+sqrt%283%29%2F2%2Ai


-----------------------------------------------------------------------
Answer:



So the roots of x%5E4%2B3x%5E3%2Bx%5E2%2B4 are x=-2 (with a multiplicity of 2), x=1%2F2+%2B+sqrt%283%29%2F2%2Ai and x=1%2F2+-+sqrt%283%29%2F2%2Ai