Question

(i) p(x) = 2x^3 + 3x^2 - 4x+3 , g(x) = x+3​

Answer 1

AnswEr:-

Given:-

• $\tt p(x) = 2x^3+3x^2-4x+3.$

• $\tt g(x) = x + 3.$

To Divide:-

• p(x) by g(x).

So lets divide by long division method:-

$\boxed{\begin{array}\quad\begin{tabular}{m{3.5em}cccc}&& 2x^2& -3x&+5 = quotient.\\\cline{3-6}\multicolumn{2}{l}{x + 3\big)}&2x^3& + 3x^2& - 4x & + 3\\&& 2x^3&+6x^2&&\\&&-&- &&\\ \cline{3-6}&&&-3x^2& - 4x& + 3\\&&&-3x^2&-9x &\\ &&&+&+&\\ \cline{4-6}&&&& 5x& + 3\\&&&&5x&+15\\ &&&&-& -\\ \cline{5-6}&&&&& - 12 = remainder. \\ \\\end{tabular}\end{array}}$

$\tt\therefore$ When given p(x) is divided by g(x) then then quotient will be 2x²-3x+5 and the Remainder will be -12.

$\rule{200}2$

VerificaTion:-

We know that:-

$\rm\mapsto Divident=(Divisor\times Quotient) +Remainder.\\ \\ \tt\mapsto 2x^3+3x^2-4x+3 = [(x+3)\times(2x^2-3x+5)]+(-12).\\ \\ \tt\mapsto 2x^3+3x^2-4x+3=(2x^3-3x^2+6x^2+5x-9x+15)-12.\\ \\ \tt\mapsto 2x^3+3x^2-4x+3=2x^3+3x^2-4x+3.\\ \\ \rm\mapsto LHS = RHS.\\ \\ \tt\mapsto{\underline{\underline{...HENCE\ \ \ VERIFIED...}}}$

$\rule{200}2$