Question

I need help on a math problem 48w^3+128w^2+4w-1 divided by 4w

Answer 1

Divide 4w to every term separately.

$\sf\dfrac{48w^3+128w^2+4w-1}{4w}$

$\sf\dfrac{48w^3}{4w}+\dfrac{128w^2}{4w}+\dfrac{4w}{4w}-\dfrac{1}{4w}$

48/4 = 12
w^3/w = w^2
So our first term is 12w^2.

128/4 = 32
w^2/w = w
So our second term is 32w.

4w/4w = 1
Anything divided by itself is 1.

-1/4w, this can't be simplified further, so this is our last term.

So we have:

$\sf 12w^2+32w+1-\dfrac{1}{4w}$