Question

Divide the quadratic polynomial $3x+1+2{{x}^{2}}$ by $x+2$ and verify division algorithm.

Answer 1

Hint: This question is all about variable division algorithm. First of all, we divide the given divisor from the given dividend till the remainder degrees become less than the degree of divisor. Then we verify the division algorithm by the formula –
Dividend $=$ Quotient $\times $ Divisor $+$ Remainder

Complete step-by-step solution
Now, let us understand the method of division:-
$\left( 2{{x}^{2}}+3x+1 \right)\div \left( x+2 \right)$
(i) First of all, we multiply divisor with a number that can cut maximum degree terms of dividend, and that will, and that written in quotient and remainder will come below. We have to subtract $2{{x}^{2}}$, so multiply with$2x$.
\[\begin{align}
  & Divisor\to \left( x+2 \right)\overset{Quotient\to 2x}{\overline{\left){2{{x}^{2}}+3x+1}\right.}}\to Dividend \\
 & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2{{x}^{2}}+4x \\
 & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\overline{~~0-x+1~~~~}~\to Remainder \\
\end{align}\]
(ii) By multiplying $2x$ with divisor, highest degree term of dividend is subtracted but remainder is$-x+1$, and degree of remainder is equal to degree of divisor. So we have to process again this procedure. Now we have to subtract remainder’s highest degree term, which is $-x$ . So we have to multiply with $-1$ .
\[\begin{align}
  & \left( x+2 \right)\overset{2x-1}{\overline{\left){2{{x}^{2}}+3x+1}\right.}} \\
 & ~~~~~~~~~~~-2{{x}^{2}}+4x \\
 & ~~~~~~~~~~~~\overline{~~~0-x+1~~~~} \\
 & ~~~~~~~~~~~~\underline{~~~~~-\left( x+2 \right)} \\
 & ~~~~~~~~~~~~~~~~~~~~~~+3 \\
\end{align}\]
So, now by multiplication with $\left( -1 \right)$, highest degree term is subtracted and remainder is $3$. Now degree of remainder is less than degree of divisor.
So, we have got
Quotient $=\left( 2x-1 \right)$
Remainder $=3$
Divisor $=\left( x+2 \right)$
Dividend $=3x+1+2{{x}^{2}}$
Now by division algorithm,
Dividend $=$ Quotient $\times $ Divisor $+$ Remainder
By taking RHS –
Quotient $\times $ Divisor $+$ Remainder
$=\left\{ \left( x+2 \right)\times \left( 2x-1 \right) \right\}+3$
$=\left\{ x\left( 2x-1 \right)+2\left( x+2 \right) \right\}+3$
$=2{{x}^{2}}-x+4x-2+3$
$=2{{x}^{2}}+3x+1$
$=LHS$
$=Dividend$

Note: (i) During the division process, after multiplying quotient with divisor when we subtract from the dividend, we have to take care of significant change because of subtraction. All terms of multiplication of quotient and divisor will have a significant change. Many students can make mistake here.
(ii) We can divide $3x+1+2{{x}^{2}}$ by $x+2$ as below:
$\dfrac{2{{x}^{2}}+3x+1}{x+2}$
(a) First we make multiple of $\left( x+2 \right)$ in numerator.
Let $2{{x}^{2}}+3x+1=Ax\left( x+2 \right)+B\left( x+2 \right)+C$
Now, compare both parts.
$A=2$
$2A+B=3$
$\Rightarrow 4+B=3$
$\Rightarrow B=-1$
And, $2B+C=1$
$\Rightarrow -2+C=1$
$\Rightarrow C=3$
So, we have got
$2{{x}^{2}}+3x+1=2x\left( x+2 \right)-1\left( x+2 \right)+3$
(b) Now,
$\dfrac{2{{x}^{2}}+3x+1}{x+2}=\dfrac{2x\left( x+2 \right)}{x+2}-1\dfrac{2x\left( x+2 \right)}{x+2}+\dfrac{3}{x+2}$
$\Rightarrow \dfrac{2{{x}^{2}}+3x+1}{x+2}=\left( 2x-1 \right)+\dfrac{3}{x+2}$
Hence, the quotient is $\left( 2x-1 \right)$ and the remainder is$3$.