1. Exam Question - Polynomials

i) Given that , find the value of A, of B and of C.

ii) Hence using the substitution , solve for , the equation

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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.

2. Re: Exam Question - Polynomials

Originally Posted by zikcau25
i) Given that , find the value of A, of B and of C.

ii) Hence using the substitution , solve for , the equation

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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.
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*}

3. 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.

4. 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.

5. 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.