Question

The polynomials \[{{x}^{3}}+2{{x}^{2}}-5ax-8\] and \[{{x}^{3}}+a{{x}^{2}}-12x-6\] when divided by \[\left( x-2 \right)\] and \[\left( x-3 \right)\] leave remainder \[p\] and \[q\] respectively. If \[q-p=10\] , find the value of ‘a’.

Answer 1

Hint: These types of problems are quite straight forward and are very easy to solve. Questions of these types can be solved quickly and correctly once we figure out the underlying concepts behind the problem. To solve it we need to have a fair amount of knowledge of polynomials and quadratic equations as well as factor theorem and division and multiplication of polynomials. In this problem, since we are given a cubic polynomial, we first find out the respective cubes of the factors that are given and then subtract it from the original polynomial to find the value of the remainder and then equate it to the given remainders given.

Complete step by step answer:
Now we start off with the solution to the given problem by finding out the values of the cubes of the factor given for the respective polynomials. For the first one the cube is,
\[{{\left( x-2 \right)}^{3}}={{x}^{3}}-8-6{{x}^{2}}+12x\] . Now subtracting this value from the original polynomial we get,
\[\begin{align}
  & \left( {{x}^{3}}+2{{x}^{2}}-5ax-8 \right)-\left( {{x}^{3}}-8-6{{x}^{2}}+12x \right) \\
 & =8{{x}^{2}}-5ax-12x \\
\end{align}\]
Now finding the value of \[8{{\left( x-2 \right)}^{2}}\] and subtracting it from the above equation we get,
 \[\begin{align}
  & 8{{x}^{2}}-5ax-12x-8{{\left( x-2 \right)}^{2}} \\
 & =20x-5ax-32 \\
\end{align}\]
Now putting \[x=2\] in this equation we find the remainder and equate it to the given remainder ‘p’,
\[\begin{align}
  & 40-10a-32=p \\
 & \Rightarrow p=8-10a \\
\end{align}\]
In the second case, we put \[x=3\] in the given polynomial to find the remainder and equate it to the given remainder ‘q’,
\[\begin{align}
  & 27+9a-36-6=q \\
 & \Rightarrow q=9a-15 \\
\end{align}\]
Now putting this values of ‘p’ and ‘q’ in the given relation we find out the value of ‘a’ as,
\[\begin{align}
  & \left( 9a-15 \right)-\left( 8-10a \right)=10 \\
 & \Rightarrow 9a+10a-23=10 \\
 & \Rightarrow 19a=33 \\
 & \Rightarrow a=\dfrac{33}{19} \\
\end{align}\]
So this is our required answer.

Note: Problems like these require a clear cut understanding of factor theorem and polynomials. We can find the required remainder from the polynomials either by dividing the polynomial with the given factor or by equating the factor to zero and finding the value of ‘x’ and putting this value in the corresponding polynomial. After doing these, we need to put these found out values of remainder in the given relation to find out the required value of ‘a’.