1. ## 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)

2. 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.

3. 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)

4. Given roots $\displaystyle x_1,x_2,...,x_n$, the quadratic can be rewritten as:

$\displaystyle \alpha(x-x_1)(x-x_2)\cdot ... \cdot (x-x_n)$ where $\displaystyle \alpha$ is some real number.

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

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

If the quadratic $\displaystyle ax^2 + bx + c$ has no real roots then it can't be factorised using real factors.

6. 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.

$\displaystyle ax^2 + bx + c = 0$

which has solutions

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

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

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

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

which has solutions:

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

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

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

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

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:

$\displaystyle dax^2 + dbx + dc$

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

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

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

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

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

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

Then the quadratic can be written:

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

If you fancy some algebra exercise, substitute $\displaystyle \frac {-b + \sqrt {b^2 - 4 ac}} {2a}$ for $\displaystyle \alpha$ and $\displaystyle \frac {-b - \sqrt {b^2 - 4 ac}} {2a}$ for $\displaystyle \beta$ , then multiply it out and prove that you get $\displaystyle ax^2 + bx + c = 0$.

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

8. 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?