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Remainder Theorem

When 2x2 -a...

Question

When $2 x^{2} - a x + 7$ and $a x^{2} + 7 x + 12$ are divided by (x-3) and (x+1) respectively, the remainder is same. Find the value of a.

a = 7

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a = 14

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a = 21

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a = 5

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Solution

The correct option is D $a = 5$
$2 x^{2} - a x + 7$ and $a x^{2} + 7 x + 12$ are divided by $(x - 3)$ and $(x + 1)$ respectively, the remainder is same.
Thus by Remainder Theorem,
$2 (3)^{2} - a (3) + 7 = a (- 1)^{2} + 7 (- 1) + 12$
$=> 18 - 3 a + 7 = a - 7 + 12$
$=> - 4 a + 25 = 5$
$=> 4 a = 20$
$=> a = 5$

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When 2x2−ax+7 and ax2+7x+12 are divided by (x-3) and (x+1) respectively, the remainder is same. Find the value of a.

The correct option is D a=52x2−ax+7 and ax2+7x+12 are divided by (x−3) and (x+1) respectively, the remainder is same.Thus by Remainder Theorem, 2(3)2−a(3)+7=a(−1)2+7(−1)+12=>18−3a+7=a−7+12=>−4a+25=5=>4a=20=>a=5

When $2 x^{2} - a x + 7$ and $a x^{2} + 7 x + 12$ are divided by (x-3) and (x+1) respectively, the remainder is same. Find the value of a.