# Math Help - values of roots

1. ## 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

2. 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 $P(x) = (x-2)(x^2-2x-2)$

3. 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?

4. 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

5. 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?

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