1. ## Solving quadratic equations by factoring

Using the "zero property" $ax^2+bx+c=0$

$4x^2=60-x$

move the $60-x$ over to the left...

$4x^2-60+x=0$

Factor out...

ok now here is where I have my problem. As seeing that the Eq has a negative and a positive the signs would be the same when factoring out no?!

$(4x-15)(x-4)$

$4x-15=0$ and $x-4=0$

$4x=+15$ and $x=4$

Thus...

$x=\frac{15}{4}$ and $x=4$

Yes?!

$4x^2-60+x = (4x- 15)(x+4)$

remember you can always check if your factorisation is correct by expanding your factors.

3. Originally Posted by bobak

$4x^2-60+x = (4x- 15)(x+4)$

remember you can always check if your factorisation is correct by expanding your factors.
.....ok how do I expand the factors, that's something I have not heard of!?

4. you just multiply out the factors.

so $(4x- 15)(x+4) = 4x \cdot x + 4x \cdot 4 -15 \cdot x - 15 \cdot 4$

5. Originally Posted by bobak
you just multiply out the factors.

so $(4x- 15)(x+4) = 4x \cdot x + 4x \cdot 4 -15 \cdot x - 15 \cdot 4$
ahhhhhh yes yes!! In the original problem I did not re-arrange the equation so that it would be in standard form!

$4x^2+x-60$

Now it makes sense!!

thanks Bobak