# factoring after finding roots with quadratic

• Oct 10th 2009, 10:08 PM
Flay
factoring after finding roots with quadratic
I'm at uni now and I dont know why this is causing me problems or why I've only just discovered this problem.

Say you have a quadratic that you want to factorise. You solve it using the quadratic equation and you have the roots and then you can find the factors.
But the quadratic formula would give you the same roots for any multiple of the quadratic. Is the best way to sort this out just to expand it and work out what the difference is between the coefficients and then just put that multiple out the front of the factors?

Thanks,
(using brothers account)
• Oct 10th 2009, 11:41 PM
Flay
I'm not sure what I've written makes sense now.. if both roots were 0 atleast, then any multiple of the quadratic would have the same roots..
i hope someone can set my thoughts straight lol.

I guess my question really is, how do you factorise a quadratic, given the roots.
• Oct 10th 2009, 11:46 PM
Quote:

Originally Posted by Flay
I'm not sure what I've written makes sense now.. if both roots were 0 atleast, then any multiple of the quadratic would have the same roots..
i hope someone can set my thoughts straight lol.

I guess my question really is, how do you factorise a quadratic, given the roots.

HI

I am not really following what you are trying to say .. sorry ..

Correct me if i am wrong . You were asking if i know the roots of this quadratic equation are lets say 3 and 4 , then what are the factors of this equation .. well in this case ,

x=3 .. the first factor would be (x-3)

x=4 .. the other one would be (x-4)
• Oct 10th 2009, 11:47 PM
Defunkt
Given roots $x_1,x_2,...,x_n$, the quadratic can be rewritten as:

$\alpha(x-x_1)(x-x_2)\cdot ... \cdot (x-x_n)$ where $\alpha$ is some real number.
• Oct 10th 2009, 11:59 PM
mr fantastic
Quote:

Originally Posted by Flay
I'm at uni now and I dont know why this is causing me problems or why I've only just discovered this problem.

Say you have a quadratic that you want to factorise. You solve it using the quadratic equation and you have the roots and then you can find the factors.
But the quadratic formula would give you the same roots for any multiple of the quadratic. Is the best way to sort this out just to expand it and work out what the difference is between the coefficients and then just put that multiple out the front of the factors?

Thanks,
(using brothers account)

If the quadratic $ax^2 + bx + c$ has two distinct real roots $x_1$ and $x_2$ then it can be factorised as $a(x - x_1)(x - x_2)$.

If the quadratic $ax^2 + bx + c$ has one repeated root $x_1$ then it can be factorised as $a(x - x_1)^2$.

If the quadratic $ax^2 + bx + c$ has no real roots then it can't be factorised using real factors.
• Oct 11th 2009, 12:01 AM
Matt Westwood
Quote:

Originally Posted by Flay
I'm not sure what I've written makes sense now.. if both roots were 0 atleast, then any multiple of the quadratic would have the same roots..
i hope someone can set my thoughts straight lol.

I guess my question really is, how do you factorise a quadratic, given the roots.

There's two things here.

You have the quadratic:

$ax^2 + bx + c = 0$

which has solutions

$x = \frac {-b \pm \sqrt{b^2 -4ac}}{2a}$.

Now suppose that same quadratic were multiplied by some non-zero real number $d$.

That is, $d (ax^2 + bx + c) = 0$.

Rewriting: $dax^2 + dbx + dc = 0$

which has solutions:

$\frac {-db \pm \sqrt {d^2 b^2 - 4 dadc}} {2da}$

which is $\frac {-db \pm d \sqrt {b^2 - 4 ac}} {2da}$

from which it is obvious that $d$ cancels out top and bottom leaving you with:

$x = \frac {-b \pm \sqrt{b^2 -4ac}}{2a}$.

So, as we see, the quadratic indeed gives you the same roots whether multiplied by a constant or not. So that's all right.

Okay, now suppose you have a quadratic (note: not an equation) in such a form:

$dax^2 + dbx + dc$

i.e. you can see there's an obvious constant factor there, $d$.

I believe that the first thing you can do is to extract that factor $d$ right out so as to get:

$d (ax^2 + bx + c)$

and then factorise (if you can) the remaining quadratic $ax^2 + bx + c$ so as to get something in the form:

$da(x - \alpha)(x - \beta)$

Right, now we've got that out of the way ... how do you factorise a quadratic, given the roots? Actually, this is really easy.

You have $ax^2 + bx + c = 0$. Suppose the roots were $\alpha$ and $\beta$.

Then the quadratic can be written:

$a (x - \alpha)(x - \beta)$.

If you fancy some algebra exercise, substitute $\frac {-b + \sqrt {b^2 - 4 ac}} {2a}$ for $\alpha$ and $\frac {-b - \sqrt {b^2 - 4 ac}} {2a}$ for $\beta$ , then multiply it out and prove that you get $ax^2 + bx + c = 0$.
• Oct 11th 2009, 12:34 AM
Flay
Ok, thanks for all your replies its much clearer now.
So whats the best way to find the constant that it is multiplied by?
• Oct 11th 2009, 12:49 AM
mr fantastic
Quote:

Originally Posted by Flay
Ok, thanks for all your replies its much clearer now.
So whats the best way to find the constant that it is multiplied by?

Did you read post #5!?
• Oct 11th 2009, 12:53 AM
Flay
Quote:

Originally Posted by mr fantastic
Did you read post #5!?

woops. Ok got it now.

thanks.