SOLUTION: please help me understand and solve the equation x^3 + 2x^2 + x + 2 = 0, if -2 is a root.

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Question 92193This question is from textbook Algebra and Trigonometry
: please help me understand and solve the equation x^3 + 2x^2 + x + 2 = 0, if -2 is a root. This question is from textbook Algebra and Trigonometry

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If -2 is a root, then -2 is a test zero. So that means we can use synthetic division

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|1212
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
-2|1212
|
1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 2)
-2|1212
|-2
1

Add -2 and 2 to get 0. Place the sum right underneath -2.
-2|1212
|-2
10

Multiply -2 by 0 and place the product (which is 0) right underneath the third coefficient (which is 1)
-2|1212
|-20
10

Add 0 and 1 to get 1. Place the sum right underneath 0.
-2|1212
|-20
101

Multiply -2 by 1 and place the product (which is -2) right underneath the fourth coefficient (which is 2)
-2|1212
|-20-2
101

Add -2 and 2 to get 0. Place the sum right underneath -2.
-2|1212
|-20-2
1010

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

Now lets look at the bottom row of coefficients:

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

x%5E2+%2B+1


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

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



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

Now lets break x%5E2+%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%2B1=0 (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like x%5E2%2B0%2Ax%2B1=0 notice a=1, b=0, and c=1)

x+=+%280+%2B-+sqrt%28+%280%29%5E2-4%2A1%2A1+%29%29%2F%282%2A1%29 Plug in a=1, b=0, and c=1



x+=+%280+%2B-+sqrt%28+0-4%2A1%2A1+%29%29%2F%282%2A1%29 Square 0 to get 0



x+=+%280+%2B-+sqrt%28+0%2B-4+%29%29%2F%282%2A1%29 Multiply -4%2A1%2A1 to get -4



x+=+%280+%2B-+sqrt%28+-4+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%280+%2B-+2%2Ai%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%280+%2B-+2%2Ai%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=0+%2B+i or x=0+-+i


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Answer:

So the polynomial has roots

x=-2, x=i and x=-i (the last two are the imaginary roots)