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)

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- Feb 7th 2013, 03:55 PMmichellederzFactoring
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 PMjakncokeRe: 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 PMmichellederzRe: Factoring
Okay, this makes sense! For some reason I couldn't find the solution.. lol. Thank you so much!

- Feb 7th 2013, 06:58 PMibduttRe: 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 PMpssingh1001Re: 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