Two polynomials $(2 x^{3} + x^{2} - 6 a x + 7)$ and $(x^{3} + 2 a x^{2} - 12 x + 4)$ are divided by $(x + 1)$ and $(x - 1)$ respectively. If $R_{1}$ and $R_{2}$ are the remainders and $2 R_{1} + 3 R_{2} = 27$ , find the value of $a$

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Solution

Let $p (x) = 2 x^{3} + x^{2} - 6 a x + 7$ and $f (x) = x^{3} + 2 a x^{2} - 12 x + 4$
$R_{1}$ is the remainder when $p (x)$ is divided by $(x + 1) ∴ R_{1} = p (- 1)$
$\Rightarrow R_{1} = 2 (- 1)^{3} + (- 1)^{2} - 6 a (- 1) + 7 = - 2 + 1 + 6 a + 7 ∴ R_{1} = 6 a + 6$
$R_{2}$ is the remainder when $f (x)$ is divided by $(x - 1) ∴ R_{2} = f (1)$
$\Rightarrow R_{2} = 1^{3} + 2 a (1)^{2} - 12 (1) + 4 = 1 + 2 a - 12 + 4 ∴ R_{2} = 2 a - 7$
Substituting the values of $R_{1}$ and $R_{2}$ in $2 R_{1} + 3 R_{2} = 27$ , we get
$2 (6 a + 6) + 3 (2 a - 7) = 27 ∴ 12 a + 12 + 6 a - 21 = 27$
$18 a - 9 = 27 ∴ 18 a = 27 + 9 = 36 ∴ a = \frac{36}{18} = 2 ∴ a = 2$

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Two polynomials (2x3+x2−6ax+7) and (x3+2ax2−12x+4) are divided by (x+1) and (x−1) respectively. If R1 and R2 are the remainders and 2R1+3R2=27, find the value of a

Two polynomials $(2 x^{3} + x^{2} - 6 a x + 7)$ and $(x^{3} + 2 a x^{2} - 12 x + 4)$ are divided by $(x + 1)$ and $(x - 1)$ respectively. If $R_{1}$ and $R_{2}$ are the remainders and $2 R_{1} + 3 R_{2} = 27$ , find the value of $a$