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.
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 $
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.