When the polynomial $x^{3} + 2 x^{2} - 5 a x - 7$ is divided by $(x - 1)$ , the remainder is A and when the polynomial $x^{3} + a x^{2} - 12 x + 16$ is divided by $(x + 2)$ , the remainder is B. Find the value of 'a' if $2 A + B = 0$ .

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Solution

It is given that when the polynomial $x^{3} + 2 x^{2} - 5 a x - 7$ is divided by (x – 1), the remainder is A.
${(1)}^{3}$ + 2 ${(1)}^{2}$ – 5a(1) – 7 = A
1 + 2 – 5a – 7 = A
– 5a – 4 = A …(i)
It is also given that when the polynomial $x^{3} + a x^{2} - 12 x + 16$ is divided by (x + 2), the remainder is B.
$x^{3} + a x^{2} - 12 x + 16$ = B
${(- 2)}^{3}$ + a ${(- 2)}^{2}$ – 12(-2) + 16 = B
-8 + 4a + 24 + 16 = B
4a + 32 = B …(ii)
It is also given that 2A + B = 0
Using (i) and (ii), we get,
2(-5a – 4) + 4a + 32 = 0
-10a – 8 + 4a + 32 = 0
-6a + 24 = 0
6a = 24
a = 4

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