# Factoring

• Feb 7th 2013, 03:55 PM
michellederz
Factoring
Pretty sure factoring would be in the algebra department... hoping so. Anyways, can anyone help me factor this equation?

3x^2-10x+3

I can't seem to figure it out. (Tmi)
• Feb 7th 2013, 04:07 PM
jakncoke
Re: Factoring
Try solving for roots using the quadratic equations. I got $\displaystyle x_1 = 3, x_2 = \frac{1}{3}$

so $\displaystyle (x-3)(x-\frac{1}{3}) = x^2 - \frac{1}{3}x - 3x + 1 = 0$
multiply left hand side by 3 to remove fraction so $\displaystyle 3*(x-\frac{1}{3})(x-3) = (3x-1)(x-3) = 3x^2-10x + 3$
• Feb 7th 2013, 04:11 PM
michellederz
Re: Factoring
Okay, this makes sense! For some reason I couldn't find the solution.. lol. Thank you so much!
• Feb 7th 2013, 06:58 PM
ibdutt
Re: Factoring
In such cases first multiply the coefficient of x^2 and constant term. In this case that is 3 *3 = 9. Find such factors of this product such that their sum equals the coefficient of x. In this question we get 9 = -1 * -9 and -1 + ( -9) = -10
Rewrite the quadratic by splitting the middle term.
3x^2 - x - 9x + 3 = x ( 3x - 1) - 3 ( 3x - 1 ) = (3x-1)(x-1) and there after it is quite simple.
• Feb 7th 2013, 09:03 PM
pssingh1001
Re: Factoring
3x^2-10x+3=0
the quadratic eq^n satisfies with x=3
then x-3=0
then we can write
i.e 3x^2-9x-x+3=0
3x(x-3) -1 (x-3)=0
i.e (x-3)(3x-1)=0
then we can write
x-3=0 & 3x-1=0
then x=3 & x=1/3