If the polynomial $6 x^{4} + 8 x^{3} - 5 x^{2} + a x + b$ is exactly divisible by the polynomial $2 x^{2} - 5$ , then find the product of the values of $a$ and $b$ .

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Solution

Consider $g (x)$ is a polynomial of $2^{n d}$ defree
$∴ 6 x^{4} + 8 x^{3} - 5 x^{2} + a x + b = (\sqrt{2} x - \sqrt{5}) (\sqrt{2} x + \sqrt{5}) \times g (x)$
Now, $R h S = 0$ when $x = \sqrt{(5 / 2)}$ & $x = - \sqrt{(5 / 2)}$
by substituting $x = \sqrt{(5 / 2)}$ ,
$6 (25 / 4) + 8 (5 \sqrt{5 / 2} \sqrt{2}) - 5 (5 / 2) + a (\sqrt{5} / \sqrt{2}) + b = 0$
$25 + 20 (\sqrt{5} / \sqrt{2}) + a (\sqrt{5} / \sqrt{2}) + b = 0$ --------(i)
by substituting $x = - \sqrt{(5 / 2)}$
$6 (25 / 4) 8 (5 \sqrt{5 / 2} \sqrt{2}) - 5 (5 / 2) - a \sqrt{5} / \sqrt{2}) + b = 0$
$25 - 20 (\sqrt{5} \sqrt{2}) - a (\sqrt{5} / \sqrt{2}) + b = 0$ .......(ii)
By $(i) + (i i)$ ,
$50 + 2 b = 0$
$b = - 25$
By $(i) - (i i)$
$40 (\sqrt{5} / \sqrt{2}) + 2 a (\sqrt{5} \sqrt{2}) = 0$
$a = - 20$
$∴$ product of values a & b $= a \times b$
$= (- 20) \times (- 25)$
$= 500$

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