how do you sovle 4x^3-2x^2+8x/2x?

Answer 1

In this question you just have (2x) below the fraction so you just can divide all of them by ' 2x '

here is our question 4x^3-2x^2+8x / 2x ;so

( 4x^3 / 2x) - (2x^2 / 2x) + (8x /2x) =
(4/2 * x^3 / x ) - (2/2 * x^2 / x) + (8/2 * x/x)=
▪ (4/2 * x^3 /x) = 2x^2
▪ (2/2 * x^2 / x) = 1 x = x
▪ (8/2 * x / x)= 4

Now just replace all of them in main formula;

2x^2 - x + 4
And its our correct answer.
Good luck.

Answer 2

You go order of operations, divide, multiply, add, and then subtract. In this case, you divide the 8x by the 2x first. The x's cancel out, so you have 8/2=4

4x^3-2x^2+4

You can't do anything else, unless you have the other side of the equation, like 4x^3-2x^2+4=0, or something like that.

Answer 3

"Solve" means find the value of x that satisfies an equation. Since there isn't an equation here, I think you mean "simplify"; I also think you mean (4x^3-2x^2+8x)/2x, in which case you should factorise the numerator by taking out the common factor 2x:

2x(2x^2 - x + 4)/2x

and then dividing both top and bottom by 2x leaves

2x^2 - x + 4

Answer 4

If you mean (4x^3-2x^2+8x)/(2x), then this equals (4x^3/2x)-(2x^2/2x)+(8x/2x) = 2x^2 - x + 4. The general solution gives x=(1 +/- sqrt(1-4*2*4))/(2*2) = (1 +/- sqrt(-31))/4. If you are not using imaginary numbers then 2x^2 -x +4 is your answer. If you are then 2x^2-x+4 = 2(x-a)(x-b) where a=(1+i*sqrt(31))/4 and b=(1-i*sqrt(31))/4.

Answer 5

First you can simplify by pulling out a 2x of the equation. That will leave the bottom to be 1. Now all you have to do is break up the top equation and solve for zero.

Answer 6

You need to clarify . . .

is it (4x^3-2x^2+8x)/2x OR

4x^3-2x^2+(8x/2x)

Answer 7

4x^3-2x^2=2x
8x/2x=4x
2x+4x

Answer 8

i got 2x^2-x+4
but im only in geometry, algebra 2 next year lol

Answer 9

(4x^3 - 2x^2 + 8x)/(2x)

(2x(2x^2 - x + 4))/(2x)

ANS : 2x^2 - x + 4

Answer 10

(4x^3-2x^2+8x)/(2x)
= 4x^3/(2x) - 2x^2/(2x) + 8x/(2x)
= 2x^2 - x + 4