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c4 (probably an easy question)

I need to express (2x^3 + x^2 - 8x - 4) / (2x^3 - 3x^2 - 2x) In the form p + q/x

I did some polynomial division, got the p value as 1 then (-4x^2 + 6x + 4) / (2x^3 - 3x^2 - 2x) take 2 out of brackets with the numerator and you get 2(-2x^2 + 3x + 2) which gives me -2/x


So my final answer was 1 + -2/x I checked the mark scheme and the answer was 1+2/x....

I just can't see why the q value is positive and not negative, any help much appreciated.
Original post by Student20
I need to express (2x^3 + x^2 - 8x - 4) / (2x^3 - 3x^2 - 2x) In the form p + q/x

I did some polynomial division, got the p value as 1 then (-4x^2 + 6x + 4) / (2x^3 - 3x^2 - 2x) take 2 out of brackets with the numerator and you get 2(-2x^2 + 3x + 2) which gives me -2/x


So my final answer was 1 + -2/x I checked the mark scheme and the answer was 1+2/x....

I just can't see why the q value is positive and not negative, any help much appreciated.


Looks as though you've made a mistake doing the division. Post your working?
2x3+x2−8x−4=(2x3−3x2−2x)+4x2−6x−42x^3 + x^2 - 8x - 4 = (2x^3 -3x^2 -2x) + 4x^2 - 6x - 4
Original post by Student20
I need to express (2x^3 + x^2 - 8x - 4) / (2x^3 - 3x^2 - 2x) In the form p + q/x

I did some polynomial division, got the p value as 1 then (-4x^2 + 6x + 4) / (2x^3 - 3x^2 - 2x) take 2 out of brackets with the numerator and you get 2(-2x^2 + 3x + 2) which gives me -2/x


So my final answer was 1 + -2/x I checked the mark scheme and the answer was 1+2/x....

I just can't see why the q value is positive and not negative, any help much appreciated.


Your division is incorrect. The remainder should be your remainder multiplied by negative one.
How did you divide - can you take a pic of your workings?
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Original post by Indeterminate
2x3+x2−8x−4=(2x3−3x2−2x)+4x2−6x−42x^3 + x^2 - 8x - 4 = (2x^3 -3x^2 -2x) + 4x^2 - 6x - 4


Oh right that seems a lot simpler than whatever I was trying to do lol
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Original post by Mr T Pities You
Your division is incorrect. The remainder should be your remainder multiplied by negative one.
How did you divide - can you take a pic of your workings?


I was using this as an example to follow

https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-division

03:07 is basically where the division starts, I was following his example, can't take a picture unfortunately haven't got a smartphone anymore.

I basically figured 2x^3 goes into 2x^3 once, so you get 1 as the p value, then you times that one by whatever is being divided (so the numerator again) then you take that away from the denominator, so basically

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

Which got me -4x^2 + 6x + 4
Original post by Student20
I was using this as an example to follow

https://www.khanacademy.org/math/algebra2/polynomial_and_rational/dividing_polynomials/v/polynomial-division

03:07 is basically where the division starts, I was following his example, can't take a picture unfortunately haven't got a smartphone anymore.

I basically figured 2x^3 goes into 2x^3 once, so you get 1 as the p value, then you times that one by whatever is being divided (so the numerator again) then you take that away from the denominator, so basically

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

Which got me -4x^2 + 6x + 4

Okay. His method is sound (at a glance) but you subtracted the wrong way around.
When using the method in that vid, be careful with sticking brackets around what you are subtracting, and also filling in 'place value' style gaps with, for example, 0x
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Original post by Mr T Pities You
Okay. His method is sound (at a glance) but you subtracted the wrong way around.
When using the method in that vid, be careful with sticking brackets around what you are subtracting, and also filling in 'place value' style gaps with, for example, 0x



Ah crap yeah, I see what I've done, did the division the wrong way around, thanks for the help.

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