SOLUTION: Given a zero x=2, of the polynomial: x^3-4x^2+21x-34=0, find the others. Did I do this right? x(x^2-4x+21)-34=0 x(x-7)(x+3)-34=0 x= -7 x= -3 If I am off base, can you s

Question 205193: Given a zero x=2, of the polynomial: x^3-4x^2+21x-34=0, find the others.
Did I do this right?
x(x^2-4x+21)-34=0
x(x-7)(x+3)-34=0
x= -7 x= -3
If I am off base, can you show the correct way to do this problem? Thank you
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
x-2=0 since 2 is a zero.
(x^3-4x^2+21x-34)/(x-2) Long division
=x^2-2x+17
x^2-2x+17=0
x=1+4i, x=1-4i The other 2 zeros. Quadratic formula (see below)
.
Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -64 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.

In the field of imaginary numbers, the square root of -64 is + or - .

The solution is

Here's your graph:

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