Engineering Polynomial - Help needed!

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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

Reply 21

ghostwalker

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.

Reply 22

clloyd12

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!

Reply 23

clloyd12

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.