Question

of polynomial please answer​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:p(y) = {y}^{4} + {y}^{3} + {8y}^{2} + my + n$

and

$\rm :\longmapsto\:q(y) = {y}^{2} + 1$

Let us now divide p(y) by q(y) using long division, we get

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: {y}^{2} + y + 7 \: \: \:}}}\\ {{\sf{ {y}^{2} + 1}}}& {\sf{\: {y}^{4} + {y}^{3} + {8y}^{2} + my + n \:}} \\{\sf{}}&\underline{\sf{\: \: \: { - y}^{4} + \: \: \: \: \: \: \: { - y}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: {y}^{3} + {7y}^{2} + my + n \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: - {y}^{3} \: \: \: \: \: \: \: \: \: - y \: \: \: \: \: \: \: \: \: \:\:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {7y}^{2} + (m - 1)y + n\:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: { - 7y}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 7 \:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (m - 1)y \: + (n - 7) \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

Now, we have

$\rm :\longmapsto\:Remainder = (m - 1)y + (n - 7)$

Since, it is given that q(y) is a factor of p(y).

$\bf\implies \:Remainder = 0$

$\rm :\longmapsto\:(m - 1)y + (n - 7) = 0$

$\rm :\longmapsto\:m - 1 = 0 \: \: \: and \: n - 7 = 0$

$\bf\implies \:m = 1 \: \: \: or \: \: \: n = 7$

Additional Information :-

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: then$

$\red{\boxed{ \rm \: \alpha + \beta = - \frac{b}{a}}}$

and

$\red{\boxed{ \rm \: \alpha\beta = \frac{c}{a}}}$

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + b {x}^{2} + cx + d, \: then$

$\red{\boxed{ \rm \: \alpha + \beta + \gamma = - \frac{b}{a}}}$

$\red{\boxed{ \rm \: \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}}}$

$\red{\boxed{ \rm \: \alpha\beta \gamma = - \frac{d}{a}}}$