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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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	|	1	2	1	2
	|

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

-2	|	1	2	1	2
	|
		1

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

-2	|	1	2	1	2
	|		-2
		1

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

-2	|	1	2	1	2
	|		-2
		1	0

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

-2	|	1	2	1	2
	|		-2	0
		1	0

Add 0 and 1 to get 1. Place the sum right underneath 0.

-2	|	1	2	1	2
	|		-2	0
		1	0	1

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

-2	|	1	2	1	2
	|		-2	0	-2
		1	0	1

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

-2	|	1	2	1	2
	|		-2	0	-2
		1	0	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 (1,0,1) form the quotient




So

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



Basically factors to

Now lets break down further



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


Starting with the general quadratic



the general solution using the quadratic equation is:



So lets solve (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 notice , , and )

Plug in a=1, b=0, and c=1



Square 0 to get 0



Multiply to get



Combine like terms in the radicand (everything under the square root)



Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

or


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Answer:

So the polynomial has roots

, and (the last two are the imaginary roots)