Roots of a Cubic Polynomial

**adkinsjr** · October 10th 2009, 02:37 PM

The equation $x^3-6x^2+21x-26=0$ has one real root and two complex roots of the form $x_1=a+bi$ and $x_2=a-bi$ where $a,b$ are real numbers. Find a+b

I'm not sure where to begin. I don't understand how to solve these kinds of expressions. Sometimes can group terms and find a way to factor the polynomial so that the roots are obvious. I don't understand how his would work if some of the roots were complex numbers. Would the equation take the form:

$(x\pm(a+bi))(x\pm(a-bi))(x\pm c)=0$

?

**skeeter** · October 10th 2009, 03:07 PM

Originally Posted by adkinsjr

The equation $x^3-6x^2+21x-26=0$ has one real root and two complex roots of the form $x_1=a+bi$ and $x_2=a-bi$ where $a,b$ are real numbers. Find a+b

I'm not sure where to begin. I don't understand how to solve these kinds of expressions. Sometimes can group terms and find a way to factor the polynomial so that the roots are obvious. I don't understand how his would work if some of the roots were complex numbers. Would the equation take the form:

$(x\pm(a+bi))(x\pm(a-bi))(x\pm c)=0$

?

use the rational root theorem to find the single real root (it does exist) ... then use synthetic division to find the quadratic factor which can be solved to find the two complex roots.

**adkinsjr** · October 10th 2009, 04:43 PM

Originally Posted by skeeter

use the rational root theorem to find the single real root (it does exist) ... then use synthetic division to find the quadratic factor which can be solved to find the two complex roots.

I don't understand even understand how to use the theorem, nor do I understand why it works.

Rational root theorem - Wikipedia, the free encyclopedia

In the example on the wikipedia page, it seems that they just took the 2 and 3 in the equation:

to get $\frac{2}{3}$ , but where did the $\frac{1}{1}$ come from?

I was thinking about finding the point of inflection by the second derivative test to find a root. Is it true that the point of inflection of a cubic function will always be on the x-axis, and will thus give us the real root?

**skeeter** · October 10th 2009, 05:00 PM

Originally Posted by adkinsjr

I don't understand even understand how to use the theorem, nor do I understand why it works.

Rational root theorem - Wikipedia, the free encyclopedia

In the example on the wikipedia page, it seems that they just took the 2 and 3 in the equation:

to get $\frac{2}{3}$ , but where did the $\frac{1}{1}$ come from?

I was thinking about finding the point of inflection by the second derivative test to find a root. Is it true that the point of inflection of a cubic function will always be on the x-axis, and will thus give us the real root?

try this link for an explanation of how the rational root theorem works ...

The Rational Roots Test

fyi , the rational root for your equation is x = 2.

**adkinsjr** · October 10th 2009, 05:10 PM

Originally Posted by skeeter

try this link for an explanation of how the rational root theorem works ...

The Rational Roots Test

fyi , the rational root for your equation is x = 2.

$f'(x)=3x^2-12x+21$ and $f''(x)=6x-12$

So when $f''(x)=0,x=2$ .

Ok, so it seems I was correct about the inflection point on the x-axis. With this information I can easily find the complex roots.

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