Question

A polynomial \[f\left( x \right)\] with rational coefficient leaves remainder \[15\], when divided by \[\left( x-3 \right)\] and remainder \[2x+1\], when divided by \[{{\left( x-1 \right)}^{2}}\]. If \[p\] is the coefficient of \[x\] of its remainder which will come out if \[f\left( x \right)\] is divided by \[\left( x-3 \right){{\left( x-1 \right)}^{2}}\] then find \[p\].

Answer 1

Hint: In order to find \[p\], we will be expressing the given statements mathematically first. Then by applying the remainder theorem to the given remainders, we will be substituting the obtained value accordingly in the respective equations to which the remainder belongs. After solving two equations, upon solving them we obtain the coefficients of both the polynomials.

Complete step by step answer:
Now let us learn about the remainder theorem. Remainder theorem is also an Euclidean approach. When we apply this theorem, when a polynomial is divided by a factor \[x-a\] which not necessarily should be a part of the polynomial, we will be finding a polynomial smaller than this along with the remainder.
Now let's find the value of \[p\].
We are said that it leaves remainder \[15\], when divided by \[\left( x-3 \right)\]
So we get \[f\left( x \right)\] as \[f\left( x \right)=\left( x-1 \right)\left( x \right)+15\to \left( 1 \right)\]
We are also told that we get the remainder \[2x+1\] when\[f\left( x \right)\] is divided by \[{{\left( x-1 \right)}^{2}}\].
So we can also write \[f\left( x \right)\] as \[f\left( x \right)={{\left( x-1 \right)}^{2}}m\left( x \right)+2x+1\to \left( 2 \right)\]
Now let us consider that \[r\left( x \right)\] would be the remainder when \[f\left( x \right)\] is divided by \[\left( x-3 \right){{\left( x-1 \right)}^{2}}\]
So we can express \[f\left( x \right)\] as \[f\left( x \right)=\left( x-3 \right){{\left( x-1 \right)}^{2}}n\left( x \right)+r\left( x \right)\to \left( 3 \right)\]
Since the polynomial is of the third degree, we can conclude that the remainder will definitely be of a degree lesser than the polynomial i.e. let us say it as second degree
So let \[r\left( x \right)=a{{x}^{2}}+bx+c\]
Considering the \[\left( 1 \right)\] and \[\left( 3 \right)\] equations, we get
\[\begin{align}
  & f\left( 3 \right)=15 \\
 & \Rightarrow r\left( 3 \right)=9a+3b+c=15\to \left( 4 \right) \\
\end{align}\]
Now upon considering \[\left( 2 \right)\] and \[\left( 3 \right)\] equations, we get
\[\begin{align}
  & f\left( 1 \right)=3 \\
 & \Rightarrow r\left( 1 \right)=a+b+c=3\to \left( 5 \right) \\
\end{align}\]
Also, from \[\left( 2 \right)\] and \[\left( 3 \right)\] equations, we get
\[\begin{align}
  & {{f}^{'}}\left( 1 \right)=2 \\
 & \Rightarrow {{r}^{'}}\left( 1 \right)=2a+b=2\to \left( 6 \right) \\
\end{align}\]
Now let us solve the equations, \[\left( 4 \right),\left( 5 \right),\left( 6 \right)\] by elimination method.
Consider \[\left( 4 \right),\left( 5 \right)\]. Upon solving it, we get
\[\begin{align}
  & \text{9a+3b+c=15} \\
 & \,\,\,\underline{\text{ a+ b+c= 3}} \\
 & \,\,\text{8a+2b =12} \\
\end{align}\]
Now,
\[\begin{align}
  & \text{8a+2b=12} \\
 & \underline{\text{4a+2b=4}} \\
 & \text{4a} \,\,\,\,\,\,\,\,\,\,\, \text{=8} \\
\end{align}\]
We get \[a=2\]
Let us substitute \[a=2\] in \[\left( 6 \right)\], we get
\[\begin{align}
  & 2a+b=2 \\
 & \Rightarrow 2\left( 2 \right)+b=2 \\
 & \Rightarrow b=-2 \\
\end{align}\]
Now we will be substituting \[a=2\] and \[b=-2\] in equation \[\left( 5 \right)\]. We get
\[\begin{align}
  & a+b+c=3 \\
 & \Rightarrow 2-2+c=3 \\
 & \Rightarrow c=3 \\
\end{align}\]
\[\therefore \] The remainder when \[f\left( x \right)\] is divided by \[\left( x-3 \right){{\left( x-1 \right)}^{2}}\] is \[2x^2-2x+3\]
So the value of p is -2.

Note: We must always have a note that the remainder of a polynomial should always possess a lesser degree than the divisor polynomial. We use the remainder theorem in finding a factor of a polynomial as it is the easiest way of finding it.