Question

If the polynomial x4 - 6x + 26x2 – 25x + 10 is divided by another polynomial x? - 2x +k
the remainder comes out to be x + a, find k and a.

Answer 1

Answer:

The value of k and a is 5 and -5 respectively.

Step-by-step explanation:

Dividend=Divisor×Quoitent+Remainder

and reminder = x+a

On diving the above given equation, we get ,

Given the reminder is (x+a)

(4k-25+16-2k)x+[10-k(8-k)] = x+a

On comparing on both sides, we get

2k-9=1

2k=10

Therefore, k=5

Also,  

10-40+25=a

Therefore, a=-5

Answer 2

Correct Question :

If the polynomial, $\bf\ x^4 - 6x^3 + 16x^2 - 25x + 10$ is divided by another polynomial $\bf\ x^2 - 2x + k$ the remainder comes out to be $\bf\ x + a$.Find $\bf\ k \: and \: a$.

$\rule{200}2$

AnswEr :

$\normalsize\sf\ By \: division \: algorithm$

$\normalsize\dashrightarrow\sf\ Dividend = Divisor \times\ Quotient + Remainder$

$\normalsize\dashrightarrow\sf\ Dividend - Remainder = Divisor \times\ Quotient$

$\normalsize\dashrightarrow\sf\ x^4 - 6x^3 + 26x^2 - 25x + 10 - x - a = x^2 + 2x + k \times\ Quotient$

$\normalsize\dashrightarrow\sf\ x^4 - 6x^3 + 26x^2 - 26x + 10 - a = x^2 + 2x + k \times\ Quotient$

$\normalsize\sf\ Hence, \: \bf\ x^4 - 6x^3 + 26x^2 - 26x + 10 - a, \: \sf\ is \: the \\ \normalsize\sf\ Polynomial \: which \: divides \\ \normalsize\bf\ x^2 + 2x + k \: \sf\ completely[ i.e. Remainder \: is \: 0]$

$\rule{200}1$

$\boxed{\begin{array}{c | l } & \quad\bf\ x^2 - 4x + (8-k) \\ \cline{1-2} x^2 + 2x + k & \: \: x^4 - 6x^3 + 16x^2 - 26x + 10 - a \\ & \: \: x^4 - 2x^3 + kx^2\\ & - \: \: \ + \:\: \: \quad - \\ \cline{2-2} & \: \: \quad\ -4x^3 + (16 - k)^2 - 26x \\ & \: \: \quad\ -4x^3 + \quad\ 8x^2 \quad\ - 4kx\\ \cline{2-2} & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \qquad\ (8-k)x^2 - (26-4k)x + 10 - a \\ & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \qquad\ (8-k)x^2 - (16 - 2k)x + (8k - k^2) \\ & \: \: \: \: \: \: \: \: \: \: \: \qquad\ - \: \: \: \: \: \: \:\: \: \: \: \: \:\quad\ +\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\ - \\ \cline{2-2} & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \qquad\ (-10 + 2k)x + (10- a - 8k + k^2)\end{array} }$

$\rule{200}1$

$\normalsize\sf\ Now; \: Remainder = 0$

$\scriptsize\texttt{\: \: \: \: \: \: \: \dag\ P(x) \: is \: factor \: of \: g(x)}$

$\normalsize\twoheadrightarrow\sf\ (-10 + 2k)x + (10 - a - 8k + k^2) = 0$

$\normalsize\twoheadrightarrow\sf\underbrace{ (-10 + 2k)x}_{part \: 1} + \underbrace{ (10 - a - 8k + k^2)}_{part \: 2} = 0$

$\rule{100}1$

$\underline{\bigstar\:\textsf{Calculation \: with \: part \: 1}}$

$\normalsize\ : \implies\sf\ (-10 + 2k)x = 0$

$\normalsize\ : \implies\sf\ -10 + 2k = 0$

$\normalsize\ : \implies\sf\ 2k = 10$

$\normalsize\ : \implies\sf\ k = \frac{\cancel{10}}{\cancel{2}}$

$\normalsize\ : \implies\sf\ k = 5$

$\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: k \: is}{\textbf{\: 5}}}$

$\underline{\bigstar\:\textsf{Calculation \: with \: part \: 2}}$

$\normalsize\ : \implies\sf\ (10 - a - 8k + k^2) = 0$

$\scriptsize\texttt{ \: \: \: \: \: \: \dag\ put \: the \: value \: of \: k}$

$\normalsize\ : \implies\sf\ 10 - a - 8k + k^2 = 0$

$\normalsize\ : \implies\sf\ 10 - a - 8(5) + (5)^2 = 0$

$\normalsize\ : \implies\sf\ (10 - a - 40 + 25 = 0$

$\normalsize\ : \implies\sf\ 10 - a - 40 + 25 = 0$

$\normalsize\ : \implies\sf\ -a - 5 = 0$

$\normalsize\ : \implies\sf\ -a = 5$

$\normalsize\ : \implies\sf\ a = - 5$

$\therefore\:\underline{\textbf{Hence, \: the \: value \: of \: of \: a \: is}{\textbf{-5}}}$