Student 999
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Struggling on the last part, it seems to me I need to translate it somehow to remove the y^3 coefficient but can't see howName:  Screenshot 2022-07-02 at 16.20.35.png
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mqb2766
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(Original post by Student 999)
Struggling on the last part, it seems to me I need to translate it somehow to remove the y^3 coefficient but can't see howName:  Screenshot 2022-07-02 at 16.20.35.png
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Size:  93.2 KB
Its the "same" as when you complete the square of a quadratic, or remove the quadratic term from a cubic.

Think of a translation like
z-a = y
so that when you sub it in, the cubic term disappears. Its obviously related to the cubic term in a 4th power binomial.
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Student 999
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(Original post by mqb2766)
Its the "same" as when you complete the square of a quadratic, or remove the quadratic term from a cubic.

Think of a translation like
z-a = y
so that when you sub it in, the cubic term disappears. Its obviously related to a 4th power binomial.
Oh I see, I just consider the case of (y-a)^4 and match coefficients to only the 4th and 3rd degrees of y giving y=x+2 as the substitution.

Is the reason why I can do that because for a general quartic x^4 +bx^3 + cx^2 + dx +e can be rewritten as (x-a_1)^4 + f(x-a_2)^2 + a_3 so a translation will affect the x^3 term of the first bracket (x-a_1)^4 whereas the rest of the brackets do not contain a x^3 term.

If I wanted to remove the x^2 term though how would I go about that, from the identity that I came up with I would need to somehow adjust the a_1 constant to produce a quartic with x^2 coefficient of -f to cancel out the quadratic term in f(x-a_2)^2 ?
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(Original post by Student 999)
Is the reason why I can do that because for a general quartic x^4 +bx^3 + cx^2 + dx +e can be rewritten as (x-a_1)^4 + f(x-a_2)^2 + a_3 so a translation will affect the x^3 term of the first bracket (x-a_1)^4 whereas the rest of the brackets do not contain a x^3 term.

If I wanted to remove the x^2 term though how would I go about that, from the identity that I came up with I would need to somehow adjust the a_1 constant to produce a quartic with x^2 coefficient of -f to cancel out the quadratic term in f(x-a_2)^2 ?
Yes for the first point. If you have a polynomial order n, you can always remove the x^(n-1) term using a transformation like that. Its what you do when you complete the square of a quadratic to remove the linear term (transform the origin), remove the quadratic term from a cubic equation etc, ...

If you had a quartic and wanted to remove the quadratic term (so keeping the quartic, cubic, linear and constant terms) you could do something similar, just expand the transformed quartic and cubic binomials and pick the "a" which makes the quadratic term zero. I don't think this is part of the question? Note you couldn't do this on both the cubic and quadratic terms. You have one knob to turn "a", so you'll only be able to control one term being zero. Note, not worked through the corresponding equation though.
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