1. Factor the expression

can somone solve this for me so i can figure out how to do the rest i got to do my teacher used some X method to solve it she put at the bottom the middle number the multiplied the first and last and put it at the top of the x

4xto the 2nd power-5x-6

2. I can't decipher how this equation is supposed to appear.

$4x^2-5x-6$ Like this?

$(4x\pm\alpha)(x\pm\beta)$

$\alpha * \beta =-6$

$4x*\beta + x*\alpha = -5x$

3. yup

4. Originally Posted by dwsmith
I can't decipher how this equation is supposed to appear.

$4x^2-5x-6$ Like this?

$(4x\pm\alpha)(x\pm\beta)$

$\alpha * \beta =-6$

$4x*\beta + x*\alpha = -5x$
Could also be $(2x\pm\alpha)(2x\pm\beta)$ ?

5. I used a CAS system to check it. It isn't 2x.

6. Fair call, I hadn't gone that far!

can somone solve this for me so i can figure out how to do the rest i got to do my teacher used some X method to solve it she put at the bottom the middle number the multiplied the first and last and put it at the top of the x

4xto the 2nd power-5x-6
$\displaystyle 4x^2 - 5x - 6$.

Multiply your $\displaystyle a$ and $\displaystyle c$ values to give $\displaystyle 4(-6) = -24$.

This means you need to find two numbers that multiply to give $\displaystyle -24$ and add to give $\displaystyle -5$.

They are $\displaystyle -8$ and $\displaystyle 3$.

So you split the middle term $\displaystyle -5x$ into $\displaystyle -8x + 3x$.

$\displaystyle 4x^2 - 5x - 6 = 4x^2 - 8x + 3x - 6$

$\displaystyle = 4x(x - 2) + 3(x - 2)$

$\displaystyle = (x - 2)(4x + 3)$.