SOLUTION: I have a problem that asks me to find the one real zero and two complex zeros. I just do not know where to even begin and our book has no examples of it. Here is the problem: 2x^3+

Question 96495: I have a problem that asks me to find the one real zero and two complex zeros. I just do not know where to even begin and our book has no examples of it. Here is the problem: 2x^3+17x^2+120x+225 Thank you so much!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'm not sure how your teacher wants you to find the first root, but we'll use a graphing calculator to find the first root.

Using the root find feature, we find one root at

Let's simplify this expression using synthetic division

Start with the given expression

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -5/2

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.

-5/2	\|	2	17	120	225
	\|

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

-5/2	\|	2	17	120	225
	\|
		2

Multiply -5/2 by 2 and place the product (which is -5) right underneath the second coefficient (which is 17)

-5/2	\|	2	17	120	225
	\|		-5
		2

Add -5 and 17 to get 12. Place the sum right underneath -5.

-5/2	\|	2	17	120	225
	\|		-5
		2	12

Multiply -5/2 by 12 and place the product (which is -30) right underneath the third coefficient (which is 120)

-5/2	\|	2	17	120	225
	\|		-5	-30
		2	12

Add -30 and 120 to get 90. Place the sum right underneath -30.

-5/2	\|	2	17	120	225
	\|		-5	-30
		2	12	90

Multiply -5/2 by 90 and place the product (which is -225) right underneath the fourth coefficient (which is 225)

-5/2	\|	2	17	120	225
	\|		-5	-30	-225
		2	12	90

Add -225 and 225 to get 0. Place the sum right underneath -225.

-5/2	\|	2	17	120	225
	\|		-5	-30	-225
		2	12	90	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 (2,12,90) form the quotient

Notice in the denominator , the x term has a coefficient of 2, so we need to divide the quotient by 2 like this:

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 ( notice , , and )

Plug in a=1, b=6, and c=45

Square 6 to get 36

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

So that means the polynomial has roots of

(which is real), and (and the last two are complex)
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