If the quotient on dividing $2 x^{4} + x^{3} - 14 x^{2} - 19 x + 6$ by $2 x + 1$ is $x^{3} + a x^{2} - b x - 6$ .
Find the values of a and b, also the remainder

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Solution

Let $p (x) = 2 x^{4} + x^{3} - 14 x^{2} - 19 x + 6$ .

Given that the divisor is $2 x + 1$ .

Now we get the value of x from

$2 x + 1 = 0$ . Then $x = - \frac{1}{2}$

$∴$ The zero of the divisor is $x = - \frac{1}{2}$ .
So, $2 x^{4} + x^{3} - 14 x^{2} - 19 x + 6 = (x + \frac{1}{2}) {2 x^{3} - 14 x - 12} + 12$

$= (2 x + 1) \frac{1}{2} (2 x^{3} - 14 x - 12) + 12$

Thus, the quotient is $\frac{1}{2} (2 x^{2} - 14 x - 12) = x^{3} - 7 x - 6$ and the remainder is 12.
But, given quotient is $x^{3} + a x^{2} - b x - 6$ . Comparing this with the quotient obtained we get, a = 0 and b = 7.
Thus, a=0, b=7, and the remainder is 12.

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