# how to do a long division for this cube expression

• Sep 26th 2006, 06:17 PM
shenton
how to do a long division for this cube expression
(4x^3 + x^2 + 8x + 4) / (x^2 + 2)

The long division as follows:

4x
_______________________
(4x^3 + x^2 + 8x + 4)
4x^3 + 8x
________
??

Due to truncation of text, the (x^2 + 2) is not shown in the long division diagram.

The workings:
4x (x^2 + 2) = 4x^3 + 8x

I'm stuck with the second step. What about the ^2 exponent?

The answer key given is 4x + 1 Remainder 2

• Sep 26th 2006, 06:30 PM
shenton
Let me try this:

For:
4x (x^2 + 2) = 4x^3 + 8x

Is it ok to append a zero to get the answer?

As follows:

4x^3 + x^2 + 8x + 4
4x^3 + 0 + 8x
__________________
x^2 + 4

Next Step:

1 (x^2 + 2) = x^2 + 2

x^2 + 4
x^2 + 2
_______
2

This approach of adding a zero to the long division allows me to get 4x+1 remainder 2. Is that right?
• Sep 26th 2006, 08:36 PM
CaptainBlack
Quote:

Originally Posted by shenton
Let me try this:

For:
4x (x^2 + 2) = 4x^3 + 8x

Is it ok to append a zero to get the answer?

As follows:

4x^3 + x^2 + 8x + 4
4x^3 + 0 + 8x
__________________
x^2 + 4

Next Step:

1 (x^2 + 2) = x^2 + 2

x^2 + 4
x^2 + 2
_______
2

This approach of adding a zero to the long division allows me to get 4x+1 remainder 2. Is that right?

Looks OK to me. You could of course check that

4x^3 + x^2 + 8x + 4= (4x+1)(x^2+2)+2

yourself to confirm that your answer is correct.

RonL
• Sep 27th 2006, 04:46 AM
Soroban
Hello, shenton!

Let me give it a try . . .

Quote:

(4x³ + x² + 8x + 4) ÷ (x² + 2)
Code:

```                 4x    +  1         --------------------   x² + 2 ) 4x³ + x² + 8x + 4           4x³  +    8x           ---------------                 x²  +    4                 x²  +    2                 -------------                           2```
• Sep 29th 2006, 09:01 AM
shenton
Thanks, Soroban for showing the workings.