# values of roots

• Sep 19th 2009, 11:10 AM
mark
values of roots
hi, the question i've got is:

the polynomial $\displaystyle P(x) = x^3 - 4x^2 + 2x + 4$ has (x - 2) as a factor

find the exact values of the roots of the equation

could someone show me step by step how you would arrive at the answer please? thankyou
• Sep 19th 2009, 11:31 AM
e^(i*pi)
Quote:

Originally Posted by mark
hi, the question i've got is:

the polynomial $\displaystyle P(x) = x^3 - 4x^2 + 2x + 4$ has (x - 2) as a factor

find the exact values of the roots of the equation

could someone show me step by step how you would arrive at the answer please? thankyou

$\displaystyle P(x) = (x-2)(ax^2 + bx + c)$

Where a,b and c are constants. You can find a, b and c by comparing coefficients

Spoiler:
.These will be of the form P(x) = (x-2)(ax^2+bx+c)

$\displaystyle x^3$ : $\displaystyle 1 = a$

$\displaystyle x^2$ : $\displaystyle -4 = -2a+b \rightarrow b = -2$

$\displaystyle x^0$ (constant) : $\displaystyle 4 = -2c \rightarrow c = -2$

Therefore $\displaystyle P(x) = (x-2)(x^2-2x-2)$
• Sep 19th 2009, 11:36 AM
mark
i already did that bit, the answer in the book is $\displaystyle x =2$, $\displaystyle x = 1 + \sqrt{3}$ and $\displaystyle x = 1 - \sqrt{3}$

any ideas on how to arrive at that answer?
• Sep 19th 2009, 11:49 AM
e^(i*pi)
Quote:

Originally Posted by mark
i already did that bit, the answer in the book is $\displaystyle x =2$, $\displaystyle x = 1 + \sqrt{3}$ and $\displaystyle x = 1 - \sqrt{3}$

any ideas on how to arrive at that answer?

The roots are where it equals 0 at.

$\displaystyle (x-2)(x^2-2x-2) = 0$

So either $\displaystyle x-2=0$ or $\displaystyle x^2-2x-2=0$

The book's answers are the solutions to those equations
• Sep 19th 2009, 12:04 PM
mark
ok thanks, how would you get to $\displaystyle x = 1 + \sqrt{3}$ and $\displaystyle x = 1 - \sqrt{3}$ from $\displaystyle (x^2 - 2x -2)$ though? what formula would you use?
• Sep 19th 2009, 12:34 PM
e^(i*pi)
Quote:

Originally Posted by mark
ok thanks, how would you get to $\displaystyle x = 1 + \sqrt{3}$ and $\displaystyle x = 1 - \sqrt{3}$ from $\displaystyle (x^2 - 2x -2)$ though? what formula would you use?

For $\displaystyle ax^2+bx+c=0$:
$\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$