Question

 Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:
\[f\left( x \right)=4{{x}^{3}}+8{{x}^{2}}+8x+7,g\left( x \right)=2{{x}^{2}}-x+1\]

Answer 1

Hint: In this question, we first need to write the relation between dividend, divisor, quotient and remainder using the division algorithm formula. Then we need to assume some linear equation as quotient and some k as remainder and substitute them in the division algorithm formula and then solve it further to get those constant values assumed.

Complete step by step answer:
Division Algorithm states that any positive integer called dividend when divided with other positive integer called divisor gives quotient and remainder.
The relation between these can be expressed as:
\[\text{Dividend}=\text{Divisor}\times \text{quotient}+\text{remainder}\]
Now, form the given question we have,
Dividend as \[f\left( x \right)=4{{x}^{3}}+8{{x}^{2}}+8x+7\] and divisor as \[g\left( x \right)=2{{x}^{2}}-x+1\]
Let us now represent the quotient as q and remainder as r.
Now, as the dividend is a cubic polynomial and the divisor is a quadratic polynomial so the quotient should be a linear polynomial as it should be multiplied with a divisor.
Let as assume the quotient and remainder as follows:
\[q=ax+b,r=k\]
Now, let us substitute all these four values in the division algorithm formula.
\[f\left( x \right)=4{{x}^{3}}+8{{x}^{2}}+8x+7,g\left( x \right)=2{{x}^{2}}-x+1,q=ax+b,r=k\]
Now, from the division algorithm formula we get,
\[\Rightarrow f\left( x \right)=g\left( x \right)\times q+r\]
Now, let us substitute the corresponding values in the above equation
\[\Rightarrow 4{{x}^{3}}+8{{x}^{2}}+8x+7=\left( ax+b\times 2{{x}^{2}}-x+1 \right)+k\]
Now, let us multiply accordingly and simplify it further.
\[\Rightarrow 4{{x}^{3}}+8{{x}^{2}}+8x+7=2a{{x}^{3}}-a{{x}^{2}}+ax+2b{{x}^{2}}-bx+b+k\]
Now, let us rewrite the above equations by writing the like terms together.
\[\Rightarrow 4{{x}^{3}}+8{{x}^{2}}+8x+7=2a{{x}^{3}}+\left( 2b-a \right){{x}^{2}}+\left( a-b \right)x+b+k\]
Now, let us compare the coefficients on both the sides to get the value of the constants.
Let us compare the coefficient of \[{{x}^{3}}\] then we get,
\[\Rightarrow 4=2a\]
Now, let us divide with 2 on both sides
\[\begin{align}
  & \Rightarrow \dfrac{4}{2}=a \\
 & \therefore a=2 \\
\end{align}\]
Now, on comparing the coefficient of \[{{x}^{2}}\]on both sides we get,
\[\Rightarrow 2b-a=8\]
Now, on substituting the value of a in the above equation we get,
\[\Rightarrow 2b-2=8\]
Now, on rearranging the terms we get,
\[\Rightarrow 2b=10\]
Now, on dividing with 2 on both sides and further simplification we get,
\[\begin{align}
  & \Rightarrow b=\dfrac{10}{2} \\
 & \therefore b=5 \\
\end{align}\]
Now, on comparing the constant terms on both sides we get,
\[\Rightarrow b+k=7\]
Now, on substituting the value of b in the above equation we get,
\[\Rightarrow 5+k=7\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow k=7-5 \\
 & \therefore k=2 \\
\end{align}\]
Now, from the quotient we assumed on substituting the respective constant values we get,
\[\begin{align}
  & \Rightarrow q=ax+b \\
 & \therefore q=2x+5 \\
\end{align}\]
Now, we get the remainder as
\[\begin{align}
  & \Rightarrow r=k \\
 & \therefore r=2 \\
\end{align}\]
Hence, the quotient is \[2x+5\] and the remainder is 2.

Note:
It is important to note that while assuming the quotient we need to assume it in such a way that on multiplying it with the divisor the degree should be equal to the dividend. If we assume any other quotient which does not satisfy the degree condition then the result will be incorrect.
While equating the coefficients it is important to equate the corresponding coefficients because equating the coefficients incorrectly or neglecting any of the terms changes the corresponding value of the constant and so the final result.