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# Any One Do A2 Maths C4??

the question is

given that f(x) = 2xcubed + 9xsqaured + 10x + 3

part a is easy just says show that -3 is a root of F(x), if you sub -3 in as x you get 0.

but part b says

Express 10/f(x) as partial fractions

now when theres a quadractic as the denominator thats fine i can split them up, but i dont know how to factorise that cubic equation to get 3 seperate fractions, any help?? been stuck on this all day.

given that f(x) = 2xcubed + 9xsqaured + 10x + 3

part a is easy just says show that -3 is a root of F(x), if you sub -3 in as x you get 0.

but part b says

Express 10/f(x) as partial fractions

now when theres a quadractic as the denominator thats fine i can split them up, but i dont know how to factorise that cubic equation to get 3 seperate fractions, any help?? been stuck on this all day.

I only do AS but I'm guessing that would work.

Banned

and

Express 10/f(x) as partial fractions

Are you sure its 10 divided by the f(x) don't usually see that sort of question.

you said -3 is a root of x, right?

so immediately you can factorise to become (x+3)(whatever)

Use long division or other to get the "whatever". It will DEFINATELY work becos -3 will flush in perfectly -its a root.

EDIT: Ahh someone above said it already

Good luck finding the answer.

Original Poster

Banned

so using the factor -3?

if you do this you got it yeah?

Beat you to it

Original Poster

i ended up with after findind a.b.c

1/x+3 + 8/2x+1 - 5/x+1

rep left for the help

Banned

Its a factor of X not a root i think

No problem :thumbsup:

Original Poster

its defo a root, question says ahow that -3 is a root of f(x)

lol being nitpicky now arn't we :-D

I say root because if you draw the graph, it will intersect y=0 at 3 places - they are called roots.

Or we could call them your way, namely a factor of f(x) because it goes in with no remainders, like 2 is a factor of 4.

Banned

Oh my bad, haha silly me

Least you got it done