Factor:
9x^2 + 0x - 64 = 0

Question

Factor:
9x^2 + 0x - 64 = 0

anonymous · Answer

does it really say $0x$??

anonymous · Answer

The original problem is just 9x^2 - 64 but my teacher told us to put +0x in between to make it easier.

anonymous · Answer

in any case 
$$9x^2-64$$ is the difference of two square 
$$a^2-b^2=(a+b)(a-b)$$ use 
$$a=3x,b=4$$

anonymous · Answer

really???? 
i would like to meet your teacher

anonymous · Answer

Lol okay, be my guest.

anonymous · Answer

is it clear what you need to do?

anonymous · Answer

no.. o.o Uh the way i get it is like  example : (x - 9) (x - 4) = 0

anonymous · Answer

I need to find an answer like that

anonymous · Answer

yes but in this case you have "the difference of two squares" meaning you have 
$$a^2-b^2$$ which factors as 
$$a^2-b^2=(a+b)(a-b)$$

lukecrayonz · Answer

Satellite, honestly my teacher said to put that to, don't ask me why.. I never bothered

lukecrayonz · Answer

TI-89 After geometry, I got tired of factoring.. haha

anonymous · Answer

omg you don't get it, every problem i solve to get ex. (x-3) has x^2 in it..

lukecrayonz · Answer

Mira, do you know how to distribute..?

anonymous · Answer

i will write the answer 
$$9x^2-64=(3x+4)(3x-4)$$
 because $9x^2$ is the square of $3x$ and $64$ is the square of $4$

anonymous · Answer

OH I GET IT THANKKSSS :D

anonymous · Answer

got that? we can do another like it 
$$16x^2-4=(4x+2)(4x-2)$$

anonymous · Answer

$$x^2-y^2=(x+y)(x-y)$$ and so on

lukecrayonz · Answer

Satellite, fan me really quickly? Thanks