Engineering Polynomial - Help needed! Watch

clloyd12
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#21
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#21
(Original post by joostan)
It's 2y^3
As such, when you equate the coefficients of y^3 there needs to be 2 on each side.
Sorry it took a while, trying to get my head around it!

Does that make:

 P(y) = (2y - 3)(y^2 + ay + b)

Therefore:

 P(y) = 2y^3 + a2y^2 + b2y - 3y^2 - 3ay - 3b

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ghostwalker
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#22
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#22
(Original post by clloyd12)
Sorry it took a while, trying to get my head around it!

Does that make:

 P(y) = (2y - 3)(y^2 + ay + b)

Therefore:

 P(y) = 2y^3 + a2y^2 + b2y - 3y^2 - 3ay - 3b

?
Yep.

It is convention and easier to read if you put the constant part at the front of each term, so 2ay^2, rather than a2y^2.

Butr what you have put is correct.
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clloyd12
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#23
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#23
(Original post by ghostwalker)
Yep.

It is convention and easier to read if you put the constant part at the front of each term, so 2ay^2, rather than a2y^2.

Butr what you have put is correct.
Awesome thanks! i'll plug this into the formula and see if I hit the right answer!
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clloyd12
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#24
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(Original post by ghostwalker)
Yep.

It is convention and easier to read if you put the constant part at the front of each term, so 2ay^2, rather than a2y^2.

Butr what you have put is correct.
Oh daym! I got it right My head honestly hurts! I just looked at the last question:

 2x^3 + 3x^2 - 7x - 5 = 0

Given

 x = - 5/2

I think this is too much maths for my day off lol
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ghostwalker
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#25
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#25
(Original post by clloyd12)
Oh daym! I got it right
:cool:


 2x^3 + 3x^2 - 7x - 5 = 0

Given

 x = - 5/2

I think this is too much maths for my day off lol
Similar method as last time, but watch your signs.
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clloyd12
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#26
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#26
(Original post by ghostwalker)
:cool:


Similar method as last time, but watch your signs.

:cool: Yeaaa it worked! Thanks so much for your help!
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