Question

on dividing 3x^2 - 2x^2 + 5x + 5 by the polynomial p(x) , the quotient and remainder are x^2 - x + 2 and -7 respectively. find p(x)?​

Answer 1

*Question:-

on dividing 3x³ - 2x² + 5x - 5 by the polynomial p(x) , the quotient and remainder are x² - x + 2 and -7 respectively. find p(x)?

*Answer-

Kindly see the attachment for your answer

Answer 2

Appropriate Question

On dividing 3x³ - 2x² + 5x - 5 by p(x), the quotient and remainder are x² - x + 2 and - 7 respectively. Find p(x).

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

Let assume that

$\rm :\longmapsto\:f(x) = {3x}^{3} - {2x}^{2} + 5x - 5$

Now

It is given that when f(x) is divided by p(x), the quotient q(x) and remainder r(x) is

$\rm :\longmapsto\:q(x) = {x}^{2} - x + 2$

and

$\rm :\longmapsto\:r(x) = - \: 7$

We know,

By Euclid Division Algorithm,

$\rm :\longmapsto\:f(x) = p(x) \times q(x) + r(x)$

On substituting the values, we get

$\rm :\longmapsto\:{3x}^{3} - {2x}^{2} + 5x - 5 = ( {x}^{2} - x + 2)q(x) - 7$

$\rm :\longmapsto\:{3x}^{3} - {2x}^{2} + 5x - 5 + 7= ( {x}^{2} - x + 2)q(x)$

$\rm :\longmapsto\:{3x}^{3} - {2x}^{2} + 5x +2= ( {x}^{2} - x + 2)q(x)$

$\rm :\longmapsto\:q(x) = \dfrac{{3x}^{3} - {2x}^{2} + 5x +2}{{x}^{2} - x + 2}$

So, Using Long Division Method,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: \: \: \: {3x} \: + \: 1\: \: \: \: \:\:}}}\\ {{\sf{ {x}^{2} - x + 2}}}& {\sf{\: {3x}^{3} - {2x}^{2} +5x +2\:}} \\{\sf{}}&\underline{\sf{\: - {3x}^{3} + {3x}^{2} - 6x \: \: \:\: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \:\: {x}^{2} - x + 2\:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: { - x}^{2} + x - 2\:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\bf\implies \:q(x) = 3x + 1$