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+

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> 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+      Log On


   

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) About Me  (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 x=-5%2F2




Let's simplify this expression using synthetic division


Start with the given expression %282x%5E3+%2B+17x%5E2+%2B+120x+%2B+225%29%2F%282x%2B5%29

First lets find our test zero:

2x%2B5=0 Set the denominator 2x%2B5 equal to zero

x=-5%2F2 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|217120225
|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
-5/2|217120225
|
2

Multiply -5/2 by 2 and place the product (which is -5) right underneath the second coefficient (which is 17)
-5/2|217120225
|-5
2

Add -5 and 17 to get 12. Place the sum right underneath -5.
-5/2|217120225
|-5
212

Multiply -5/2 by 12 and place the product (which is -30) right underneath the third coefficient (which is 120)
-5/2|217120225
|-5-30
212

Add -30 and 120 to get 90. Place the sum right underneath -30.
-5/2|217120225
|-5-30
21290

Multiply -5/2 by 90 and place the product (which is -225) right underneath the fourth coefficient (which is 225)
-5/2|217120225
|-5-30-225
21290

Add -225 and 225 to get 0. Place the sum right underneath -225.
-5/2|217120225
|-5-30-225
212900

Since the last column adds to zero, we have a remainder of zero. This means 2x%2B5 is a factor of 2x%5E3+%2B+17x%5E2+%2B+120x+%2B+225

Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,12,90) form the quotient

2x%5E2+%2B+12x+%2B+90


Notice in the denominator 2x%2B5, the x term has a coefficient of 2, so we need to divide the quotient by 2 like this:
%282x%5E2+%2B+12x+%2B+90%29%2F2=x%5E2+%2B+6x+%2B+45

So %282x%5E3+%2B+17x%5E2+%2B+120x+%2B+225%29%2F%282x%2B5%29=x%5E2+%2B+6x+%2B+45

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



Basically 2x%5E3+%2B+17x%5E2+%2B+120x+%2B+225 factors to %282x%2B5%29%28x%5E2+%2B+6x+%2B+45%29

Now lets break x%5E2+%2B+6x+%2B+45 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%2B6%2Ax%2B45=0 ( notice a=1, b=6, and c=45)

x+=+%28-6+%2B-+sqrt%28+%286%29%5E2-4%2A1%2A45+%29%29%2F%282%2A1%29 Plug in a=1, b=6, and c=45



x+=+%28-6+%2B-+sqrt%28+36-4%2A1%2A45+%29%29%2F%282%2A1%29 Square 6 to get 36



x+=+%28-6+%2B-+sqrt%28+36%2B-180+%29%29%2F%282%2A1%29 Multiply -4%2A45%2A1 to get -180



x+=+%28-6+%2B-+sqrt%28+-144+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-6+%2B-+12%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+=+%28-6+%2B-+12%2Ai%29%2F%282%29 Multiply 2 and 1 to get 2



After simplifying, the quadratic has roots of

x=-3+%2B+6%2Ai or x=-3+-+6%2Ai




So that means the polynomial has roots of


x=-5%2F2 (which is real), x=-3+%2B+6%2Ai and x=-3+-+6%2Ai (and the last two are complex)