1. ## solving this

its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

$\displaystyle -6x^2+13x-5$

so far i got $\displaystyle (3x- ?)(2x+?)$
i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the $\displaystyle 6x^2$ by x it by constant 5.

2. Originally Posted by johnsy123
its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

$\displaystyle -6x^2+13x-5$

so far i got $\displaystyle (3x- ?)(2x+?)$
i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the $\displaystyle 6x^2$ by x it by constant 5.
1. You can't solve a term. I assume that you want to factor the given term ... ? If so:

$\displaystyle -6x^2+13x-5 = -\left(6x^2-13x+5 \right)$

Since $\displaystyle 13 = 2 \cdot 5 + 3$ you can factor this term into:

$\displaystyle -\left(6x^2-13x+5 \right) = -(2x-1)(3x-5)$

3. Originally Posted by johnsy123
its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

$\displaystyle -6x^2+13x-5$

so far i got $\displaystyle (3x- ?)(2x+?)$
i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the $\displaystyle 6x^2$ by x it by constant 5.
Assuming the expression is = 0...

$\displaystyle -6x^2 + 13x - 5 = 0$

$\displaystyle -(6x^2 - 13x + 5) = 0$

You need to find two numbers that multiply to become $\displaystyle 6 \cdot 5 = 30$ that also add to be $\displaystyle -13$.

I think you'll find that the numbers are $\displaystyle -3$ and $\displaystyle -10$.

So break up the middle term then factor by grouping:

$\displaystyle -(6x^2 - 3x - 10x + 5) = 0$

$\displaystyle -[3x(2x - 1) - 5(2x - 1)] = 0$

$\displaystyle -(2x - 1)(3x - 5) = 0$.

Now use the Null Factor Law:

$\displaystyle 2x - 1 = 0$ or $\displaystyle 3x - 5 = 0$.

So $\displaystyle x = \frac{1}{2}$ or $\displaystyle x = \frac{5}{3}$.

4. Originally Posted by johnsy123
its pre straight forward, ive got an idea on how to do it, but i keep getting the wrong answer.

$\displaystyle -6x^2+13x-5$

so far i got $\displaystyle (3x- ?)(2x+?)$
i know how to solve it, but im getting the wrong numbers in the question marks above. ive got the idea of getting rid of the $\displaystyle 6x^2$ by x it by constant 5.
You can always use the quadratic formula, where

a = -6

b = 13

c = -5