1. ## quadratic equations

hi, could anyone help me understand how to do quadratic equations. i've got the equation $3x^2 - 7x + 2$ and it needs to be factorised. i was hoping someone could show me a systematic way of doing it. i read somewhere that you need to find 2 numbers that add together to make -7 and multiply together to make 2 but i don't think that applies here. the answer is $(x - 2) (3x - 1)$ . could someone show me a simple way of figuring it out so i don't have to resort to trial and error.

thanks, Mark

2. Originally Posted by mark
hi, could anyone help me understand how to do quadratic equations. i've got the equation $3x^2 - 7x + 2$ and it needs to be factorised. i was hoping someone could show me a systematic way of doing it. i read somewhere that you need to find 2 numbers that add together to make -7 and multiply together to make 2 but i don't think that applies here. the answer is $(x - 2) (3x - 1)$ . could someone show me a simple way of figuring it out so i don't have to resort to trial and error.

thanks, Mark
Hi mark,

Here's a technique you can use on this type of quadratic expression where the leading coefficient is not 1.

$3x^2-7x+2$

First, multiply the leading coefficient by the constant:

$3 \times 2 = 6$

Think of two numbers whose product is 6 and sum is -7.

You should come up with -1 and -6 pretty quickly.

Replace the middle term of your original quadratic with these two factors.

$3x^2-1x-6x+2$

Group the terms and factor:

$(3x-1x)-(6x-2)$

$x(3x-1)-2(3x-1)$

$(x-2)(3x-1)$

3. hi masters (or anyone reading this),

i don't think i understand what happens when you get to group the terms and factor.

$(3x^2 - 1x) - (6x - 2)$ i understand that part just shown, but then why do you go on to $x(3x - 1x) -2(3x - 1)$ ?

and how do you get $(x - 2) (3x - 1)$ after all that?

thanks

4. (3x^2-1x)-(6x-2)
taking x common from (3x^2-1x) and 2 from (6x-2) we get
[x(3x-1)-2(3x-1)]
now from the whole term take (3x-1) common so we get
(3x-1)(x-2)
this is done for FACTORIZATION

5. from the whole term take (3x-1) common? so you just remove it completely? why do you do that? (if thats what you do)

6. You could simply re-expand the new term to see that it is the same:

$(3x-1)(x-2) = 3x^2 -7x +2$
$x(3x-1) -2 (3x-1) = 3x^2 - x -6x + 2 = 3x^2 -7x +2$

7. thanks