For this question

(64x^3 + 0x + 8) / (4x +2)

I got 16x^2 -8X + 6

Am I right

2. Anybody?

3. Originally Posted by cleanwithit06
For this question

(64x^3 + 0x + 8) / (4x +2)

I got 16x^2 -8X + 6

Am I right
I just did the calculations and got the answer to be $\displaystyle 16x^2 - 8x +4$. I'll post up my method, wait a few minutes.

EDIT: I don't know how to use latex to do long division so I'll show how my answer is correct by opening up the brackets to obtain the original equation. My original method was using long division and divided $\displaystyle 64x^3 + 8$ by $\displaystyle 4x + 2$.

$\displaystyle (4x + 2)(16x^2 - 8x + 4)$
$\displaystyle = 64 x^3 - 32x^2 + 16x + 32x^2 - 16x + 8$
$\displaystyle = 64x^3 + 8$

4. Originally Posted by Air
I just did the calculations and got the answer to be $\displaystyle 16x^2 - 8x +4$. I'll post up my method, wait a few minutes.
I got that too.

5. Originally Posted by Air
I just did the calculations and got the answer to be $\displaystyle 16x^2 - 8x +4$. I'll post up my method, wait a few minutes.
Thankyou.

6. Ok I keep getting 16x^2 - 8x + 6

I don't know what the crap I'm doing wrong.

7. Sorry if I haven't explained clear enough.

8. $\displaystyle \frac{{64x^3 + 8}} {{4x + 2}} = \frac{{8\left( {8x^3 + 1} \right)}} {{2\left( {2x + 1} \right)}} =$$\displaystyle \frac{{4\left( {8x^3 + 1} \right)}} {{\left( {2x + 1} \right)}} = \frac{{4\left( {2x + 1} \right)\left( {4x^2 - 2x + 1} \right)}} {{\left( {2x + 1} \right)}} = 4\left( {4x^2 - 2x + 1} \right) = 16x^2 - 8x + 4$

9. Originally Posted by cleanwithit06
Hey. The only step I don't get with your method is the 4th one. It says -32 -32 comes out to being 16x or something, how is that so?

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