Question

on
dividing 4x^ 3 - 8x² + 8x +1 by a polynomial g(x) the quotient and remainder are (2x^2-3x+2) and (x+3) respectively. Find g(x).​

Answer 1

Answer:

we know that

dividend= division×quotient + remainder

so, by putting the value in their correct u will get the answer

Answer 2

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

Given that

☆ Polynomial,

$\rm :\longmapsto\:f(x) = {4x}^{3} - {8x}^{2} + 8x + 1$

☆ Quotient,

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

☆ Remainder,

$\rm :\longmapsto\:r(x) = x + 3$

☆ Now to find the divisor, g(x)

☆ By Division Algorithm, we know

☆ Dividend = Divisor × Quotient + Remainder

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

$\rm :\longmapsto\:{4x}^{3} - {8x}^{2} + 8x + 1 = ( {2x}^{2} - 3x + 2) \times g(x) + x + 3$

$\rm :\longmapsto\:{4x}^{3} - {8x}^{2} + 8x + 1 - x - 3 = ( {2x}^{2} - 3x + 2) \times g(x)$

$\rm :\longmapsto\:{4x}^{3} - {8x}^{2} + 7x - 2 = ( {2x}^{2} - 3x + 2) \times g(x)$

$\bf\implies \:g(x) = \dfrac{ {4x}^{3} - {8x}^{2} + 7x - 2 }{ {2x}^{2} - 3x + 2 }$

☆ Using long division,

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\:2x - 1\:\:}}}\\ {\underline{\sf{ {2x}^{2} - 3x + 4}}}& {\sf{\: {4x}^{3}-{8x}^{2} + 7x - 2 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  -  {4x}^{3}+6{x}^{2} - 4x \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \: \: - 2{x}^{2} + 3x - 2 \: \:  \:  \:  \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \:  \:  \: \: \: \: \: \: \: \: \: \:  2{x}^{2}  - 3x  + 2\:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\: \:  \: \:  \:  \:  \:  \:  \:0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

Hence,

$\bf\implies \:g(x) = 2x - 1$