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# SMO 2015 Q10 Round 1

When a polynomial $$f(x)$$ is divided by $$(x - 1)$$and $$(x + 5)$$, the remainders are -6 and 6 respectively. Let $$r(x)$$ be the remainder when $$f(x)$$ is divided by $$x^2 +4x - 5$$. Find the value of $$r(-2)$$.

(A) 0 (B) 1 (c) 2 (D) 3 (E) 5

How does one do this sort of question?

I'm not experienced with polynomials, so it may seem a simple question to you but not to me. But please help! I'm a learner too :)

Note by Timothy Wan
1 year, 4 months ago

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What is $$r$$ in $$x^2+4r-5$$? · 1 year, 4 months ago

It's an error. Just corrected the note. Sorry! · 1 year, 4 months ago

Have you read the remainder factor theorem? If so, how do you think this could apply? Staff · 1 year, 4 months ago

yeah you are right, but here, we would apply more of Euclids Division Algorithm for polynomials. Won't we? (Which essentially constitutes the proof of remainder theorem) · 1 year, 4 months ago

There are many equivalent ways of expressing the ideas involved in this question. I was pointing out one possible approach, which I think should be thought of given the phrasing of the question. Staff · 1 year, 4 months ago

Since the divisor is of degree 2, $$r(x)$$ is linear and in the form $$Ax+B$$.

$$f(1)=A(1)+B=A+B=-6$$.

Also $$f(-5)=A(-5)+B=-5A+B=6$$.

Solving the simultaneous equations, $$A=-2,B=-4$$

$$r(x)=-2x-4$$

$$\therefore r(-2)=-2(-2)-4=0$$

Is this a correct solution? · 1 year, 4 months ago

You got confused between $$f(x)$$ and $$r(x)$$. · 1 year, 4 months ago