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Let $R 1$ and $R 2$ are the remainders when the polynomials $f (x) = 4 x 3 + 3 x 2 − 12 a x − 5$ and $g (x) = 2 x 3 + a x 2 − 6 x + 2$ are divided by $x − 1 a n d x + 2$ respectively. If $3 R 1 + R 2 + 28 = 0$ find the value of ' a '.

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answer is A.

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Detailed Solution

It is given,

f (x) = 4 x 3 + 3 x 2 − 12 a x − 5

g (x) = 2 x 3 + a x 2 − 6 x + 2

Understand that according to question,

R 1

is the remainder when the polynomial

f x = 4 x 3 + 3 x 2 − 12 a x − 5

is divided by

x − 1

.
Therefore, calculate the value of

R 1

413 + 312 − 12 a 1 − 5 = R 1 ⇒ R 1 = 4 + 3 − 12 a − 5

⇒ R 1 = 2 − 12 a

Understand that according to question,

R 2

is the remainder when the polynomial

g x = 2 x 3 + a x 2 − 6 x + 2

is divided by

x + 2

.
Therefore, calculate the value of

R 2

2 − 2 3 + a − 2 2 − 6 − 2 + 2 = R 2 ⇒ R 2 = − 16 + 4 a + 12 + 2 ⇒ R 2 = − 2 + 4 a

Calculate the value of

a

.
Substitute the values of

R 1

and

R 2

3 R 1 + R 2 + 28 = 0

∴ 3 2 − 12 a + − 2 + 4 a + 28 = 0 ⇒ 6 − 36 a − 2 + 4 a + 28 = 0 ⇒ 32 a = 32 ⇒ a = 1

The value of

a = 1

.
Hence, the correct option is 1.

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