The polynomials $2 x^{3} - 7 x^{2} + a x - 6$ and $x^{3} - 8 x^{2} + (2 a + 1) x - 16$ leave the same remainder when divided by $x - 2$ . Find the value of ' $a$ '.

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Solution

Let $f (x) = 2 x^{3} - 7 x^{2} + a x - 6$

Put $x - 2 = 0$

$\Rightarrow x = 2$

When $f (x)$ is divided by $(x - 2)$ , remainder $= f (2)$

$∴ f (2) = 2 (2)^{3} - 7 (2)^{2} + a .2 - 6$

$= 2.8 - 7.4 + 2 a - 6$

$= 16 - 28 - 6 + 2 a$

$= 2 a - 18$

Let $g (x) = x^{3} - 8 x^{2} + (2 a + 1) x - 16$

when $g (x)$ is divided by $(x - 2)$ remainder $= g (2)$

$∴ g (2) = (2)^{3} - 8 (2)^{2} + (2 a + 1) 2 - 16$

$= 8 - 32 + 4 a + 2 - 16$

$= 4 a - 38$

By the condition we have

$f (2) = g (2)$

$2 a - 18 = 4 a - 38$

$4 a - 2 a = 38 - 18$

$2 a = 20$

$a = \frac{20}{2}$

$a = 10$

$∴$ Thus, the value of $a = 10$

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The polynomials 2x3−7x2+ax−6 and x3−8x2+(2a+1)x−16 leave the same remainder when divided by x−2. Find the value of 'a'.

The polynomials $2 x^{3} - 7 x^{2} + a x - 6$ and $x^{3} - 8 x^{2} + (2 a + 1) x - 16$ leave the same remainder when divided by $x - 2$ . Find the value of ' $a$ '.