values of roots

• Sep 19th 2009, 12:10 PM
mark
values of roots
hi, the question i've got is:

the polynomial $P(x) = x^3 - 4x^2 + 2x + 4$ 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, 12:31 PM
e^(i*pi)
Quote:

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

the polynomial $P(x) = x^3 - 4x^2 + 2x + 4$ 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

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

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)

$x^3$ : $1 = a$

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

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

Therefore $P(x) = (x-2)(x^2-2x-2)$
• Sep 19th 2009, 12:36 PM
mark
i already did that bit, the answer in the book is $x =2$, $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$

any ideas on how to arrive at that answer?
• Sep 19th 2009, 12:49 PM
e^(i*pi)
Quote:

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

any ideas on how to arrive at that answer?

The roots are where it equals 0 at.

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

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

The book's answers are the solutions to those equations
• Sep 19th 2009, 01:04 PM
mark
ok thanks, how would you get to $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$ from $(x^2 - 2x -2)$ though? what formula would you use?
• Sep 19th 2009, 01:34 PM
e^(i*pi)
Quote:

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

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