# Cubic Polynomial

• Jun 23rd 2010, 06:25 AM
rtblue
Cubic Polynomial
A Cubic Polynomial with integral coefficients has a root $(3-4i)$ and possesses the form:

$y=x^3+ax+b$ Find AB/25

Have no clue where to start with this one. Any help is appreciated.
• Jun 23rd 2010, 06:41 AM
Ackbeet
If you know one root, you should be able to factor the cubic. One way to do that is by polynomial long division. The result should look like your y.
• Jun 23rd 2010, 06:52 AM
rtblue
How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us

$x^2-6x+25$

Now where do I go from here?
• Jun 23rd 2010, 07:05 AM
p0oint
You found that well.

Now just do this (x-(3-4i))(x-(3+4i) or $((x-3)+4i)((x-3)-4i)=(x-3)^2-(4i)^2$

Regards.
• Jun 23rd 2010, 07:33 AM
Unknown008
Well, I tried using the factor theorem.

Let $f(x) = x^3 + ax + b$

Since 3-4i is a factor, f(3-4i) = 0.

So, $f(3-4i) = (3-4i)^3 + a(3-4i) + b = 0$

Expanding, you get:

$(b - 3a -117) - (44 + 4a)i = 0$

So,

$b - 3a - 117 = 0$

and

$44 + 4a = 0$

I'm not sure if that's right though (Thinking)

This gives a = -11, and b = 150

So, $\frac{ab}{25} = \frac{(-11)(150)}{25} = -66$
• Jun 23rd 2010, 11:01 AM
HallsofIvy
Quote:

Originally Posted by rtblue
How can I factor the polynomial, if the A and B values are not given? On another note, If one of the roots of the polynomial is (3-4i), then another root has to be (3+4i). multiplication of the two gives us

$x^2-6x+25$

Now where do I go from here?

Very good. Now, if "c" is the third root, then you must have $(x^2- 6x+ 25)(x- c)= x^3+ ax + b$.

$(x^2- 6x+ 25)(x- c)= x^3- 6x^2+ 25x- cx^2+ 6cx- 25c= x^3- (6+ c)x^2+ (25+ 6c)x- 25c$
and that must be equal to $x^3+ ax+ b$

The fact that there is no " $x^2$" term on the right tells you that $6+ c= 0$ so c= -6. Now, it should be easy to find a and b.