# Thread: Cubic Polynomial

1. ## Cubic Polynomial

A Cubic Polynomial with integral coefficients has a root $\displaystyle (3-4i)$ and possesses the form:

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

Have no clue where to start with this one. Any help is appreciated.

2. 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.

3. 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

$\displaystyle x^2-6x+25$

Now where do I go from here?

4. You found that well.

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

Regards.

5. Well, I tried using the factor theorem.

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

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

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

Expanding, you get:

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

So,

$\displaystyle b - 3a - 117 = 0$

and

$\displaystyle 44 + 4a = 0$

I'm not sure if that's right though

This gives a = -11, and b = 150

So, $\displaystyle \frac{ab}{25} = \frac{(-11)(150)}{25} = -66$

6. 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

$\displaystyle x^2-6x+25$

Now where do I go from here?
Very good. Now, if "c" is the third root, then you must have $\displaystyle (x^2- 6x+ 25)(x- c)= x^3+ ax + b$.

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

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