Exam Question - Polynomials

i) Given that http://www.texify.com/img/\LARGE\!(3...x^2%2BCx^4.gif, find the value of **A**, of **B** and of **C**.

ii) Hence using the substitution http://www.texify.com/img/\LARGE\!y%3Dx^2.gif, solve for http://www.texify.com/img/\LARGE\!x.gif, the equation

http://www.texify.com/img/\LARGE\!(3...)^5%3D1086.gif

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I wonder if there is a direct way to solve the problem without expansion, which is evidently not the right approach.

Re: Exam Question - Polynomials

Quote:

Originally Posted by

**zikcau25**

Why is expansion not the right approach? Expanding binomials is easy...

$\displaystyle \displaystyle \begin{align*} (a + b)^n = \sum_{k = 0}^n{{n\choose{k}} a^{n-k} b^k} \end{align*}$

Re: Exam Question - Polynomials

wow...

but I am not a Genius yet....

I am low level student..

So, will you please solve the questions in steps ... as it is required at Cambridge Sc Level please.

Re: Exam Question - Polynomials

No, I will not solve the question in steps. That's against the policy of the site. This is for YOU to do.

I suggest you read about Pascal's Triangle. The notation $\displaystyle \displaystyle \begin{align*} {n\choose{k}} \end{align*}$ stands for "row $\displaystyle \displaystyle \begin{align*} n \end{align*}$, position $\displaystyle \displaystyle \begin{align*} k \end{align*}$" in the triangle (and the 1 at the top counts as row 0). Then you can write

$\displaystyle \displaystyle \begin{align*} (a + b)^5 = {5\choose{0}}a^5 + {5\choose{1}}a^4b + {5\choose{2}}a^3b^2 + {5\choose{3}}a^2b^3 + {5\choose{4}}ab^4 + {5\choose{5}}b^5 \end{align*}$

which is the simplified version for the Binomial Expansion I showed you earlier.

Re: Exam Question - Polynomials

That a great guide.

Yes, indeed Sir, to solve Polynomials with larger powers ... the only handy way is Binomial Theorem. I need to get used to that more often.

Thanking you and Soroban a lot. :)