Let $R_{1}$ and $R_{2}$ are the remainder when the polynomials $x^{3} + 2 x^{2} - 5 a x - 7$ and $x^{3} + a x^{2} - 12 x + 6$ are divided by $x + 1$ and $x - 2$ respectively. If $2 R_{1} + R_{2} = 6$ , find the value of $a$ .

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Solution

Let $p (x) = x^{3} + 2 x^{2} - 5 a x - 7$
and $q (x) = x^{3} + a x^{2} - 12 x + 6$ be the given polynomials,
Now, $R_{1} =$ Remainder when p(x) is divided by x+1.
$\Rightarrow R_{1} = p (- 1)$
$\Rightarrow R_{1} = (- 1)^{3} + 2 (- 1)^{2} - 5 a (- 1) - 7 [∵ p (x) = x^{2} + 2 x^{2} - 5 a x - 7]$
$\Rightarrow R_{1} = - 1 + 2 + 5 a - 7$
$\Rightarrow R_{1} = 5 a - 6$
And $R_{2} =$ Remainder when q(x) is divided by x-2
$\Rightarrow R_{1} = q (2)$
$\Rightarrow R_{2} = (2)^{3} + a \times 2^{2} - 12 \times 2 + 6 [∵ q (x) = x^{2} + a x^{2} - 12 x - 6]$
$\Rightarrow R_{2} = 8 + 4 a - 24 + 6$
$\Rightarrow R_{2} = 4 a - 10$
Substituting the values of $R_{1}$ and $R_{2}$ in $2 R_{1} + R_{2} = 6$ , we get
$\Rightarrow 2 (5 a - 6) + (4 a - 10) = 6$
$\Rightarrow 10 a - 12 + 4 a - 10 = 6$
$\Rightarrow 14 a - 22 = 6$
$\Rightarrow 14 a - 28 = 0$
$\Rightarrow a = 2$

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Let R1 and R2 are the remainder when the polynomials x3+2x2−5ax−7 and x3+ax2−12x+6 are divided by x+1 and x−2 respectively. If 2R1+R2=6, find the value of a.

Let $R_{1}$ and $R_{2}$ are the remainder when the polynomials $x^{3} + 2 x^{2} - 5 a x - 7$ and $x^{3} + a x^{2} - 12 x + 6$ are divided by $x + 1$ and $x - 2$ respectively. If $2 R_{1} + R_{2} = 6$ , find the value of $a$ .