Question

if the remainder on division of (x^3+2x^2+kx+3)by (x-3)is 21 find the quotient.

answer is (x^2+5x+6)
but i want the steps​

Answer 1

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

Given polynomial is

$\rm :\longmapsto\: {x}^{3} + {2x}^{2} + kx + 3$

Let assume that

$\rm :\longmapsto\: f(x) = {x}^{3} + {2x}^{2} + kx + 3$

Now, it is given that

$\rm :\longmapsto\:f(x) \: leaves \: remainder \: 21 \: when \: divided \: by \: x - 3$

We know,

Remainder Theorem states that if a polynomial f(x) is divided by linear polynomial (x - a), then remainder is f(a).

So, By Remainder Theorem

$\rm :\longmapsto\:f(3) = 21$

$\rm :\longmapsto\: {(3)}^{3} + {2(3)}^{2} + k(3) + 3 = 21$

$\rm :\longmapsto\:27 + 18 + 3k + 3 = 21$

$\rm :\longmapsto\:48 + 3k = 21$

$\rm :\longmapsto\:3k = 21 - 48$

$\rm :\longmapsto\:3k = - 27$

$\bf :\longmapsto\:k = -9$

So, given polynomial can be rewritten as

$\rm :\longmapsto\: f(x) = {x}^{3} + {2x}^{2} - 9x + 3$

Now, using long division,

$\: \red{\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + 5x + 6\:\:}}}\\ {\underline{\sf{x - 3}}}& {\sf{\: {x}^{3} + {2x}^{2} - 9x + 3 \:\:}} \\{\sf{}}& \underline{\sf{ \:  -  {x}^{3} + 3 {x}^{2}   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 5{x}^{2} - 9x + 3  \:  \:  \:  \:   \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \: - 5{x}^{2} + 15x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \:  \:  \:  \:  \:  \:  \:  \:  6x + 3  \:\:}} \\{\sf{}}& \underline{\sf{\: \:  \:  \:  \:  \:  \:  \:  \:  \ - 6x + 18\:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 21\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}}$

So,

$\bf :\longmapsto\:Quotient = {x}^{2} + 5x + 6$