# Math Help - solving this

1. ## solving this

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

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

so far i got $(3x- ?)(2x+?)$
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 $6x^2$ 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.

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

so far i got $(3x- ?)(2x+?)$
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 $6x^2$ 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:

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

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

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

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.

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

so far i got $(3x- ?)(2x+?)$
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 $6x^2$ by x it by constant 5.
Assuming the expression is = 0...

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

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

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

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

So break up the middle term then factor by grouping:

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

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

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

Now use the Null Factor Law:

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

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

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.

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

so far i got $(3x- ?)(2x+?)$
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 $6x^2$ by x it by constant 5.
You can always use the quadratic formula, where

a = -6

b = 13

c = -5