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Answer 1

Hello friends!!

Here is your answer :

7.

$p(x) = 2 {x}^{4} + {x}^{3} - 8 {x}^{2} - x + 6$

$g(x) = 2x - 3$



$zero \: \: of \: \: g(x) = \frac{3}{2}$


When we divide p(x) by g (x) using remainder theorem, the remainder will be
$p( \frac{3}{2} ) = 2 {( \frac{3}{2} )}^{4} + {( \frac{3}{2}) }^{3} - 8 {( \frac{3}{2} )}^{2} - ( \frac{3}{2} ) - 6$


$= 2 \times \frac{81}{16} + \frac{27}{8} - 8 \times \frac{9}{4} - \frac{3}{2} - 6$


$= \frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} - 6$


$= \frac{81 + 27 - 144 - 12 - 48}{8}$


$= \frac{108 - 204}{8}$


$= \frac{ - 99}{8}$


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9.
$p(x) = 7 { x}^{2} - 4 \sqrt{2} x - 6$

$g(x) = x - \sqrt{2}$


$zero \: \: of \: \: g(x) = \sqrt{2}$


When we divide p(x) by g(x) using remainder theorem, the remainder will be


$p( \sqrt{2} ) = 7 {( \sqrt{2}) }^{2} - 4 \sqrt{2} ( \sqrt{2} ) - 6$


$= 7 \times 2 - 4 \times 2 - 6$


$= 14 - 8 - 6$


$= 0$


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Hope it helps you...

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