Question

The polynomials ${{x}^{3}}+2{{x}^{2}}-5ax-8$ and ${{x}^{3}}+a{{x}^{2}}-12x-6$ when divided by $x-2$ and $x-3$ gives the remainders p and q respectively. If $p-q=10$, find the value of a.

Answer 1

Hint: To solve this question, we should use the concept of division in polynomials. We can write a polynomial $p\left( x \right)$ whose degree is n and a divisor $d\left( x \right)$ whose degree is x and a remainder r. We can write a relation that $p\left( x \right)=q\left( x \right)d\left( x \right)+r$. We can write that the polynomial $q\left( x \right)$ will have a degree $n-x$. In our question, we should take polynomials $p\left( x \right)$ as the given third degree polynomials and $d\left( x \right)$ as the given first degree terms. By assuming a second degree polynomial and writing the equation and by substituting $x=2$ or $x=3$ depending on the $d\left( x \right)$, we get the values of the remainder in terms of a. After that, we use $p-q=10$ and get the value of a.

Complete step-by-step solution:
For a given polynomial $p\left( x \right)$ of degree n and another polynomial $d\left( x \right)$ with degree $x$p\left( x \right)=q\left( x \right)d\left( x \right)+r$
Where the degree of $q\left( x \right)$ will be $n-x$ and r is the remainder.
Let us consider the $p\left( x \right)={{x}^{3}}+2{{x}^{2}}-5ax-8$, $d\left( x \right)=x-2$ and $q\left( x \right)=a{{x}^{2}}+bx+c$. We get
$\Rightarrow{{x}^{3}}+2{{x}^{2}}-5ax-8=\left( x-2 \right)\left( a{{x}^{2}}+bx+c \right)+{{r}_{1}}\to \left( 1 \right)$
Let us substitute $x=2$ in the above equation. We get
$\begin{align}
  & {{2}^{3}}+2\times {{2}^{2}}-5a\times 2-8=\left( 2-2 \right)\left( a.{{2}^{2}}+b\times 2+c \right)+{{r}_{1}} \\
 &\Rightarrow 8+8-10a-8=0\times \left( 4a+2b+c \right)+{{r}_{1}} \\
 &\Rightarrow {{r}_{1}}=8-10a \\
\end{align}$
We are given that the remainder is p. So, we can write that ${{r}_{1}}=p$.
$p={{r}_{1}}=8-10a$
Let us consider the $p\left( x \right)={{x}^{3}}+a{{x}^{2}}-12x-6$, $d\left( x \right)=x-3$ and $q\left( x \right)=a{{x}^{2}}+bx+c$. We get
$\Rightarrow{{x}^{3}}+a{{x}^{2}}-12x-6=\left( x-3 \right)\left( a{{x}^{2}}+bx+c \right)+{{r}_{2}}\to \left( 2 \right)$
Let us substitute $x=3$ in the above equation. We get
$\begin{align}
  & {{3}^{3}}+a\times {{3}^{2}}-12\times 3-6=\left( 3-3 \right)\left( a.{{3}^{2}}+b\times 3+c \right)+{{r}_{2}} \\
 &\Rightarrow 27+9a-36-6={{r}_{2}} \\
 &\Rightarrow {{r}_{2}}=9a-15 \\
\end{align}$
We are given that the remainder is q. So, we can write that ${{r}_{2}}=q$.
$q={{r}_{2}}=9a-15$
We are given that $p-q=10$. Substituting the values of p and q, we get
$\begin{align}
  & \left( 8-10a \right)-\left( 9a-15 \right)=10 \\
 &\Rightarrow 8-10a-9a+15=10 \\
 &\Rightarrow -19a=10-23 \\
 &\Rightarrow -19a=-13 \\
 &\Rightarrow a=\dfrac{13}{19} \\
\end{align}$
$\therefore $The value of a is $a=\dfrac{13}{19}$.

Note: Some students try to solve the equations and get the values of the quotient and the remainder from the equations -1 and 2. We can also get the answer using this method but it is a lengthy process and we are just asked about the remainders and finding them is enough for the question. There is a theorem called remainder theorem which states that the remainder when a polynomial function in x $f\left( x \right)$ is divided by $x-a$, we get a remainder r which is equal to $r=f\left( a \right)$. We used this theorem indirectly to solve the question.