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# Maths help watch

1. So i have to sketch a grAPH and i've found the y intercept by making x=0. I'm now trying to find the x intercepts, but its in the format of (
2(x+2)^3 -5(x+2)^2 -6(x+2) + 9. Normally I would set each bracket to be 0, but I don't know. I tried setting it to 9 and -9. I just can't do it. I got up desmos and saw the roots and have no idea how to get them.
2. (Original post by mariejuana)
So i have to sketch a grAPH and i've found the y intercept by making x=0. I'm now trying to find the x intercepts, but its in the format of (
2(x+2)^3 -5(x+2)^2 -6(x+2) + 9. Normally I would set each bracket to be 0, but I don't know. I tried setting it to 9 and -9. I just can't do it. I got up desmos and saw the roots and have no idea how to get them.
2(x+2)^3 -5(x+2)^2 -6(x+2) = - 9
set each bracket equal -9 and maybe it would work.
3. (Original post by mariejuana)
So i have to sketch a grAPH and i've found the y intercept by making x=0. I'm now trying to find the x intercepts, but its in the format of (
2(x+2)^3 -5(x+2)^2 -6(x+2) + 9. Normally I would set each bracket to be 0, but I don't know. I tried setting it to 9 and -9. I just can't do it. I got up desmos and saw the roots and have no idea how to get them.
Think about putting a new variable, say y = (x + 2) and see if it factorises ...
4. (Original post by Muttley79)
Think about putting a new variable, say y = (x + 2) and see if it factorises ...
Why would I do that for?
5. (Original post by mariejuana)
Why would I do that for?
It's easier to find a factor- I found one quickly.

Then replace 'y' by 'x + 2' to get the roots.

You could keep the (x + 2) and try to factorise this cubic but that's what you need to do to find intercepts on x- axis.
6. (Original post by Brudor2000)
2(x+2)^3 -5(x+2)^2 -6(x+2) = - 9
set each bracket equal -9 and maybe it would work.
Incorrect I'm afraid.
7. (Original post by mariejuana)
So i have to sketch a grAPH and i've found the y intercept by making x=0. I'm now trying to find the x intercepts, but its in the format of (
2(x+2)^3 -5(x+2)^2 -6(x+2) + 9. Normally I would set each bracket to be 0, but I don't know. I tried setting it to 9 and -9. I just can't do it. I got up desmos and saw the roots and have no idea how to get them.
you could expand the whole equation to get 2x^3 + 7x^2 -2x - 7
then factorise the equation to get x=-3.5 and x=1 and x=-1.
8. (Original post by Brudor2000)
you could expand the whole equation to get 2x^3 + 7x^2 -2x - 7
then factorise the equation to get x=-3.5 and x=1 and x=-1.
That will take FAR more time than my approach ... there's a reason it's all expressed in terms of (x + 2).
9. (Original post by Muttley79)
That will take FAR more time than my approach ... there's a reason it's all expressed in terms of (x + 2).
you are right. making y = (x+2) and then factorizing is easier then the method i suggested
10. (Original post by Brudor2000)
you are right. making y = (x+2) and then factorizing is easier then the method i suggested
The only 'danger' is remembering to change solution for y back to those for x.

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Updated: February 17, 2018
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