Question

using long division method check whether the second polynomial is a factor of the first polynomial ...plz Help
show me with Proces​

Answer 1

i think this would help

Answer 2

$\underline{\underline{\Huge{\textbf{Concept Behind:}}}}$

$\sf\textsf{Let f(x) be First polynomial and}\\\sf\textsf{ g(x) be Second Polynomial}$

$\sf\textsf{To find whether one polynomial is a Factor}\\ \sf\textsf{of other polynomial}$

$\sf\textsf{Here we use } \textbf{Polynomial Long Division Method}$

$\sf\textbf{In the division, g(x) is considered as dividend}\\\sf\textbf{and f(x) is considered as divisor.}$

$\bigstar \: \: \boxed{\mathbf{Rem\left(\dfrac{f(x)}{g(x)}\right)}\: \begin{cases}=\: 0; & \textsf{g(x) is a factor of f(x)}\\\atop\\\neq \: 0;& \textsf{g(x) is not a factor of f(x)}\end{cases}}$

$\mathsf{Here,\: Rem\left(\dfrac{f(x)}{g(x)}\right)\: implies\: Remainder\: when}\\\sf\textsf{f(x) is divided by g(x)}$



$\underline{\LARGE{\textbf{Steps for Polynomial Long Division:}}}$

$\blacktriangleright\: \: \sf\textsf{Consider both polynomials f(x) and g(x) and}\\\sf\textsf{arrange their indices in descending order}$

$\textbf{\textdagger} \: \: \sf\textsf{It may be noted that the polynomial}\\\sf\textsf{may not have all the indices from highest} \\ \sf \textsf{index when arranged in descending order.} \\ \\ \sf \textsf{For such missing terms, 0 can be considered}$

$\blacktriangleright\: \: \sf\textsf{To get first term of the quotient, f(x) is divided by g(x).}$

$\blacktriangleright\: \: \sf\textsf{This first term of Quotient is multiplied with }\\\sf \textsf{g(x) and the terms obtained are subtracted from f(x), } \\ \sf \textsf{then the remainder is to be noted.}$

$\textbf{\textdagger}\: \: \sf\textsf{Here, some terms may get cancelled.}$

$\blacktriangleright\: \: \sf\textsf{The Remainder of result of the Subtraction} \\\sf\textsf{(if terms not get cancelled) and other terms from f(x)}\\\sf\textsf{are brought down to continue the division. }$

$\blacktriangleright\: \: \sf\textsf{The same process is now continued to get } \\ \sf \textsf{the remaining terms of the quotient.}$

$\blacktriangleright\: \: \sf \textsf{The division is continued till the remainder}\\\sf \textsf{is either 0 or the division can no longer be done}\\\sf \textsf{(i.e., if the highest index of g(x) becomes greater}\\\sf \textsf{than the highest index of reminder.)}$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\mathsf{Let\: f(x) = x^3-4x^2-3x+5 \: (first \: polynomial) } \\ \mathsf{and \: g(x) = x-3 \: (second \: polynomial) }$

$\boxed{\begin{minipage}{6 cm}\quad \begin{array}{m{3.5em}cccc}& &x^2&-x&-6\\\cline{1-6}\multicolumn{2}{l}{x-3\big)}&x^3&-4x^2&-3x&+5\\&& x^3&-3x^2&\downarrow& \bigm|\\\cline{3-4}&&&-x^2&-3x&\bigm|\\&&&-x^2&+3x &\downarrow\\\cline{4-5}&&&& -6x&+5\\&&&&-6x&+18\\\cline{5-6}&&&&&-13\\\end{array}\end{minipage}}\quad\boxed{\begin{minipage}{3cm}\sf I\: \bullet\: \dfrac{x^3}{x}\; \; \, =\: \: x^2\\&\\\sf II\: \bullet\: \dfrac{-x^2}{x}=-x\\&\\\sf III\: \bullet \dfrac{-6x}{x}=-6\end{minipage}}$

$\sf\textsf{Here, the Rem \neq 0}\\\implies\: \sf\textsf{g(x) is not factor of f(x)}$



$\therefore$

$\blacksquare\: \: \Large{\underline{\large{\textbf{(x-3) is not a Factor of (x^3-4x^2-3x+5) }}}}$