Question

Who wants to HELP ME with this one ? Simplify completely: 8x^3+6x^2-4x/-2x

a.−4x^2 − 3x + 2
b.−4x^2 + 3x + 2
c.4x^4 + 3x^3 − 2x^2
d.4x^2 + 6x^2 − 4x

Answer 1

Did you mean this? this is what you typed...
\(8x^3+6x^2-\frac{4x}{-2x}\)

Answer 2

yes but all that is over -2x

Answer 3

wait like this? \[\frac{8x^2+6x^2-4x}{-2x}\]

Answer 4

yes but its 8x^3 not 8x^2 ..

Answer 5

ok yeah I saw that typo. First of all you need to remember your order of operations because what you asked was not that. remember division comes before addition so if you want all of that to be divided by the \(-2x\) you need to put brackets around the top part.

Answer 6

anyways \[\frac{8x^3+6x^2-4x}{-2x}\] is the same thing as \[\frac{8x^3}{-2x}+\frac{6x^2}{-2x}-\frac{4x}{-2x}\] so I hope you can do those three divisions

Answer 7

notice how each fraction is just a constant times a power of x? so you can further think about breaking it down like this (this is just the first of three)
\[\frac{8}{-2}*\frac{x^3}{x}\]
now the first one should be simple, for the second just remember that for exponents with the same base you add the exponents if you multiply them and subtract the exponents if you divide. so this becomes \[-4*x^{3-1}=-4x^2\]

Answer 8

ok. so i did 8x^2=64 + 6x^2=36 + 4x am i doing it right ?

Answer 9

just do that for the other two. I have to go to my class now but make sure you do that. The answer you should get is A.

Answer 10

no you didn't do it right

Answer 11

okay ill try, thanks for helping me ! :)

Answer 12

you have squared 8, but you aren't squaring 8 you are squaring the variable and multipliying that by 8. and no worries!