Question

The polynomial \[a{{x}^{3}}+3{{x}^{2}}-3\] and \[2{{x}^{3}}-5x+a\] are divided by \[\left( x-4 \right)\] , then the remainders are \[{{R}_{1}},{{R}_{2}}\] respectively. If \[2{{R}_{1}}-{{R}_{2}}=0\] find the value of \['a'\] .

Answer 1

Hint: For solving this problem we use the remainder theorem that is if a polynomial \[f\left( x \right)\] is divided by \[\left( x-k \right)\] then the remainder is given as \[f\left( k \right)\] . We use this remainder theorem to both the polynomials and use the given condition to solve for \['a'\] .

Complete step by step answer:
Let us assume that the given first polynomial as
 \[\Rightarrow {{f}_{1}}\left( x \right)=a{{x}^{3}}+3{{x}^{2}}-3\]
We are given that this polynomial is divided by \[\left( x-4 \right)\] and leaves remainder as \[{{R}_{1}}\] .
We know that the remainder theorem states that if a polynomial \[f\left( x \right)\] is divided by \[\left( x-k \right)\] then the remainder is given as \[f\left( k \right)\] .
By using this remainder theorem we can write
 \[\Rightarrow {{R}_{1}}={{f}_{1}}\left( 4 \right)\]
Now, by substituting \[x=4\] in the function we get
 \[\begin{align}
  & \Rightarrow {{R}_{1}}=a{{\left( 4 \right)}^{3}}+3{{\left( 4 \right)}^{2}}-3 \\
 & \Rightarrow {{R}_{1}}=64a+48-3 \\
 & \Rightarrow {{R}_{1}}=64a+45 \\
\end{align}\]
Now, let us consider the second polynomial given as
 \[\Rightarrow {{f}_{2}}\left( x \right)=2{{x}^{3}}-5x+a\]
We are given that this polynomial is divided by \[\left( x-4 \right)\] and leaves remainder as \[{{R}_{2}}\] .
We know that the remainder theorem states that if a polynomial \[f\left( x \right)\] is divided by \[\left( x-k \right)\] then the remainder is given as \[f\left( k \right)\] .
By using this remainder theorem we can write
 \[\Rightarrow {{R}_{2}}={{f}_{2}}\left( 4 \right)\]
Now, by substituting \[x=4\] in the function we get
 \[\begin{align}
  & \Rightarrow {{R}_{2}}=2{{\left( 4 \right)}^{3}}-5\left( 4 \right)+a \\
 & \Rightarrow {{R}_{2}}=128-20+a \\
 & \Rightarrow {{R}_{2}}=a+108 \\
\end{align}\]
We are given one condition that is \[2{{R}_{1}}-{{R}_{2}}=0\]
By substituting the values of \[{{R}_{1}},{{R}_{2}}\] in above equation we get
 \[\begin{align}
  & \Rightarrow 2\left( 64a+45 \right)-\left( a+108 \right)=0 \\
 & \Rightarrow 128a+90-a-108=0 \\
 & \Rightarrow 127a=18 \\
 & \Rightarrow a=\dfrac{18}{127} \\
\end{align}\]

Therefore the value of \['a'\] is \[\dfrac{18}{127}\] .

Note: Students may make mistakes in taking the remainder theorem. The theorem states that if a polynomial \[f\left( x \right)\] is divided by \[\left( x-k \right)\] then the remainder is given as \[f\left( k \right)\] . This theorem is applicable only if the divisor is a linear polynomial. If a polynomial \[f\left( x \right)\] is divided by a polynomial other than linear polynomial then this theorem doesn’t hold. Students may apply this theorem for those polynomials also. So, this theorem is to be taken care of.