How to solve this question on a quadratic polynomial in a short period of time? The remainders from two divisions are given.

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Answer 1 · 2018-03-18T09:02:58

Answer is (3)

Using remainder theorem we have

$P (1) = 4$ $P(1)=4$ and $P (- 1) = 6$ $P(-1)=6$

Further $P (0) = 1$ $P(0)=1$ means $P (x) = a x^{2} + b x + 1$ $P(x)=ax^2+bx+1$

as $P (1) = 4$ $P(1)=4$ , $a + b = 3$ $a+b=3$ and as $P (- 1) = 6$ $P(-1)=6$ , $a - b = 5$ $a-b=5$

Hence $a = 4$ $a=4$ and $b = - 1$ $b=-1$

and $P (2) = 4 \cdot 2^{2} - 1 \cdot 2 + 1 = 15$ $P(2)=4*2^2-1*2+1=15$

and $P (- 2) = 4 \cdot {(- 2)}^{2} - 1 \cdot (- 2) + 1 = 19$ $P(-2)=4*(-2)^2-1*(-2)+1=19$

Hence, answer is (3)