Question

Find the divisor $g(x)$, when the polynomial $p(x)=4{{x}^{3}}+2{{x}^{2}}-10x+2$ is divided by $g(x)$ and the quotient and remainder obtained are $2{{x}^{2}}+4x+1$ and 5 respectively.

Answer 1

Hint: Here, we have to find the divisor $g(x)$ by the formula: $\text{Dividend=Divisor }\!\!\times\!\!\text{ Quotient+Remainder}$
 Now, substitute all the given data into the formula and we will get $g(x)$ of the form $\dfrac{4{{x}^{3}}+2{{x}^{2}}-10x-3}{2{{x}^{2}}+4x+1}$. Now, with the help of polynomial division we can find $g(x)$.

Complete Step-by-Step solution:
Here, we are given the polynomial $p(x)=4{{x}^{3}}+2{{x}^{2}}-10x+2$, quotient is $2{{x}^{2}}+4x+1$ and remainder is 5.
Now, we have to find the divisor $g(x)$,
We know that the number which we divide is called the dividend. The number by which we divide is called the divisor. The result obtained is called the quotient. The number left over is called the remainder.
Here, $p(x)=4{{x}^{3}}+2{{x}^{2}}-10x+2$ is the dividend and the number which divides $p(x)$ is called the divisor $g(x)$, quotient is $2{{x}^{2}}+4x+1$ and remainder is 5.
We also know that:
$\text{Dividend=Divisor }\!\!\times\!\!\text{ Quotient+Remainder}$
We can substitute the quotient, dividend, divisor and remainder to the above formula. Hence we will get:
$4{{x}^{3}}+2{{x}^{2}}-10x+2=g(x)\times \left( 2{{x}^{2}}+4x+1 \right)+5$
 In the next step, take 5 to the left side, it becomes -5, we will get:
$4{{x}^{3}}+2{{x}^{2}}-10x+2-5=g(x)\left( 2{{x}^{2}}+4x+1 \right)$
Here, on the left side subtract the constant terms, we will obtain:
$4{{x}^{3}}+2{{x}^{2}}-10x-3=g(x)\left( 2{{x}^{2}}+4x+1 \right)$
Next, by cross multiplication we will get:
$\dfrac{4{{x}^{3}}+2{{x}^{2}}-10x-3}{2{{x}^{2}}+4x+1}=g(x)$
Now, we have to find the value of $g(x)$ by polynomial division. The division is as follows:
$2{{x}^{2}}+4x+1\overset{2x-3}{\overline{\left){\begin{align}
  & 4{{x}^{3}}+2{{x}^{2}}-10x-3 \\
 & 4{{x}^{3}}+8{{x}^{2}}+2x \\
 & \overline{\begin{align}
  & 0{{x}^{3}}-6{{x}^{2}}-12x-3 \\
 & \text{ }-6{{x}^{2}}-12x-3 \\
 & \text{ }\overline{0{{x}^{2}}+0x+0} \\
\end{align}} \\
\end{align}}\right.}}$
In polynomial division, first we have to divide $4{{x}^{3}}$ by $2{{x}^{2}}$, we will get $2x$ as the quotient and then multiply $2x$ by $2{{x}^{2}}+4x+1$ and put the values $4{{x}^{3}}+8{{x}^{2}}+2x$ under $4{{x}^{3}}+2{{x}^{2}}-10x+2$ and then do the subtraction, we will get $-6{{x}^{2}}-12x-3$. Now, again divide $-6{{x}^{2}}$ by $2{{x}^{2}}$, we will get the quotient -3, now, multiply -3 by $2{{x}^{2}}+4x+1$, we will get $-6{{x}^{2}}-12x-3$. Put this value under$-6{{x}^{2}}-12x-3$and do the subtraction, we will get the remainder zero.
Hence, the quotient thus obtained is $2x-3$ which is our $g(x)$.
Therefore, $g(x)=2x-3$.

Note: Here, you can verify your answer by substituting $g(x)$ in the formula $\text{Dividend=Divisor }\!\!\times\!\!\text{ Quotient+Remainder}$, $g(x)$ is the divisor, so you multiply $g(x)=2x-3$ by $2{{x}^{2}}+4x+1$ and then add 5. If you are getting $4{{x}^{3}}+2{{x}^{2}}-10x+2$then $g(x)=2x-3$ is the correct answer.