SMO 2015 Q10 Round 1

When a polynomial f(x)f(x) is divided by (x1)(x - 1) and (x+5)(x + 5), the remainders are -6 and 6 respectively. Let r(x)r(x) be the remainder when f(x)f(x) is divided by x2+4x5x^2 +4x - 5. Find the value of r(2)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
3 years, 9 months ago

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What is rr in x2+4r5x^2+4r-5?

Aditya Agarwal - 3 years, 9 months ago

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It's an error. Just corrected the note. Sorry!

Timothy Wan - 3 years, 9 months ago

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Have you read the remainder factor theorem? If so, how do you think this could apply?

Calvin Lin Staff - 3 years, 9 months ago

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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)

Aditya Agarwal - 3 years, 9 months ago

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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.

Calvin Lin Staff - 3 years, 9 months ago

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

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

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

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

r(x)=2x4r(x)=-2x-4

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

Is this a correct solution?

Timothy Wan - 3 years, 9 months ago

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You got confused between f(x)f(x) and r(x)r(x).

Anupam Nayak - 3 years, 9 months ago

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