polynomials quick question (1 Viewer)

[Managore]

Suppose x^3– 2x^2 + a ≡ (x + 2)Q(x) + 3 where Q(x) is a polynomial.
Find the value of a.



I REALLY should know this, but I have no idea. Can someone explain how to derive an answer?

 

[Danny11]

oops mistake srry.

its: a = -8x + 3

just do it by inspection, let Q(x) = (x^2 - 4x)

 

[jumb]

Minus 3 from both sides, then divide both sides by (x + 2).

i get a = 19

edit: my original answer was wrong, i did osmething naughty

 

[Danny11]

divde both sides by x+ 2 and minus 3 from teh final answer.

I ended up with -6 + a - 6 = 0

so a = 12

how can you divide both side?

 

[Managore]

I had Qx = (x^2 - 4x + 8) to cancel out the -8x, so then
a = 16 + 3 = 19 but I think it's wrong...

 

[Managore]

Ah, a friend just helped me. Substitute x= - 2 so (x+2)Qx = 0

which gives:

-8 - 8 + a = 3
a = 19

 

[Danny11]

lol. didn't know "a" was suppose to be constant.

 

[akira276]

damn man...all u have to do is sub x=-2 for both sides.....(x+2)Q(x) will cancel out and the other side will have numbers and blah bloody blah....yeh....a=19

 

[jumb]

well my way worked but yours was easier

 

[Managore]

akira, that's what I said later on But thanks anyway

 

[Bi0mic]

damn man...all u have to do is sub x=-2 for both sides.....(x+2)Q(x) will cancel out and the other side will have numbers and blah bloody blah....yeh....a=19

Yep ... always remember to take that remainder into account

 

[Managore]

Jumb, when you said divide by, you meant the annoying long devision method, right?

 

[jumb]

Yes. Yes I did.

 

[Managore]

*shudder* I hated that.. I used it in the trials twice so I could get two equations and thingy them to get x and y... bad idea

 

[Rorix]

hooray for people that understand remainder theorm