SOLUTION: I know that I must place a -0 between the ^4 and -8^2 but I believe I am doing the problem incorrect (3x^4 -8x^2 -3x-3) / (x^2-3) x^2-3 / 3x^4-0-8x^2-3x-3 becomes 3x

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Question 415631: I know that I must place a -0 between the ^4 and -8^2 but I believe I am doing the problem incorrect
(3x^4 -8x^2 -3x-3) / (x^2-3)
x^2-3 / 3x^4-0-8x^2-3x-3
becomes 3x^2 -x +0 -3/x^-3
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your dividend is 3x^4 - 8x^2 - 3x - 3

your divisor is x^2 - 3

you need to put a place holder for the x^3 term.

your new dividend becomes 3x^4 + 0x^3 - 8x^2 - 3x - 3

you are dividing x^2 - 3 into that.

a picture of the long division is shown below:

the first thing you do is divide x^2 into 3x^4 to get 3x^2

then you multiply x^2-3 by 3x^2 to get 3x^4 - 9x^2

then you subtract 3x^4 - 9x^2 from 3x^4 + 0x^3 - 8x^2 to get x^2

then you bring down the -3x and the -3 to get x^2 - 3x - 3

then you divide x^2 from (x^2-3) into x^2 from (x^2-3x - 3) to get + 1

then you multiply x^2-3 by 1 to get x^2 - 3.

then you subtract x^2 - 3 from x^2 - 3x - 3 to get -3x

since -3x exponent is less than x^2 exponent, the division stops here.

your answer is that the result of the division is equao to 3x^2 + 1 with a remainder of -3x.

to confirm your division is correct, you multiply the quotient of 3x^2 + 1 by the divisor of x^2 - 3 together and then you add the remainder back in.

x^2 - 3 times 3x^2 + 1 results in 3x^4 - 8x^2 - 3

you now add the remainder of -3x back in to get 3x^4 - 8x^2 - 3x - 3

since this is the same as your original dividend, you can be pleased to know that the division was done successfully.

as it turns out, you could have gotten away by not including the 0x^3 place holder but you don't always know that up front so it's wise to put it in and then be glad that you didn't need it rather than not put it in and find out that you did need it.