There are $x^{4} + 57 x + 15$ pens to be distributed in a class of $x^{2} + 4 x + 2$ students. Each student should get the minimum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed $(x ϵ N)$ .

9 x - 13

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9 x - 15

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9 x - 20

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9 x + 13

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Solution

The correct option is A $9 x - 13$
Number of pens to be distributed in a class $= x^{4} + 57 x + 15$
number of students $= z^{2} + 4 x + 2$
Divisor polynomial is $x^{2} + 4 a + 2$ & dividend is $x^{4} + 57 x + 15$
Dividing
$x^{2} + 4 x + 2) \frac{x^{2} - 4 x + 14}{x^{4} + 57 x + 15 \frac{x^{4} + {4 x}^{3} + 2 x^{2}}{- 4 x^{3} - 2 x^{2} + 57 x \frac{- 4 x^{3} - 16 x^{2} - 8 x}{14 x^{2} + 65 x + 15 \frac{14 x^{2} + 56 x + 28}{9 x - 13}}}}$
$∴$ each students gets $x^{2} - 4 x + 14$ pens & no. of pens left undistributed is $9 x - 13$

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There are x4+57x+15 pens to be distributed in a class of x2+4x+2 students. Each student should get the minimum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed (xϵN).

There are $x^{4} + 57 x + 15$ pens to be distributed in a class of $x^{2} + 4 x + 2$ students. Each student should get the minimum possible number of pens. Find the number of pens received by each student and the number of pens left undistributed $(x ϵ N)$ .