Question

If \[2{{x}^{3}}+a{{x}^{2}}+bx-6\] has \[\left( x-1 \right)\] as a factor and leaves remainder 2 when divided by \[\left( x-2 \right)\], find the value of a and b respectively.
(a) -8, 12
(b) 8, 12
(c) -4, 10
(d) 1, -10

Answer 1

Hint: Use the fact that if f (x) is a polynomial and is divided by \[x-\alpha \] then \[f\left( \alpha \right)\] is its remainder. So find f (1), f (2) and then equate it with 0, 2 and then solve the equations simultaneously for a, b.

Complete step-by-step answer:
In the question we are given an expression of polynomial \[2{{x}^{3}}+a{{x}^{2}}+bx-6\]. We are also told that \[\left( x-1 \right)\] is a factor of the expression and if divided by \[\left( x-2 \right)\] we get remainder 2. Now we have to find a, b.
There is a property of polynomials which I want to provide before proceeding.
If the polynomial is represented by f (x) and if it is divided by \[x-\alpha \] then it’s remainder will be \[f\left( \alpha \right)\].
Here the polynomial is given as, \[2{{x}^{3}}+a{{x}^{2}}+bx-6\] and it is represented as f (x) and as it is divided by x – 1 and leaves remainder 0. So we can write it as
\[\begin{align}
  & \Rightarrow f\left( x \right)=2{{x}^{3}}+a{{x}^{2}}+bx-6 \\
 & \Rightarrow f\left( 1 \right)=2{{\left( 1 \right)}^{3}}+a{{\left( 1 \right)}^{2}}+b\left( 1 \right)-6 \\
 & \Rightarrow f\left( 1 \right)=2+a+b-6 \\
 & \Rightarrow f\left( 1 \right)=a+b-4 \\
\end{align}\]
We know that f (1) is 0 so we get,
\[\Rightarrow a+b-4=0\]
Or, \[a+b=4\] - (i)
Now if the same polynomial is taken f (x) and is divided by x – 2, here it leaves reainder 2 so we can write it as,
\[\begin{align}
  & \Rightarrow f\left( 2 \right)=2{{\left( 2 \right)}^{3}}+a{{\left( 2 \right)}^{2}}+b\left( 2 \right)-6 \\
 & \Rightarrow f\left( 2 \right)=16+4a+2b-6 \\
 & \Rightarrow f\left( 2 \right)=4a+2b+10 \\
\end{align}\]
We know that f (2) is equal to 2 so we get,
\[\Rightarrow 4a+2b+10=2\]
Or, \[4a+2b=-8\]
Or, \[2a+b=-4\] - (ii)
Now subtracting equation (i) from equation (ii) we get,
\[\Rightarrow \left( 2a+b \right)-\left( a+b \right)=-4-4\]
Or, a = - 8
So we get the value of a is -8.
Let substitute a = - 8 in equation (i) so we get,
\[\Rightarrow -8+b=4\]
Or, b = 12
Hence the values of a and b are -8 and 12 respectively. So the correct answer is (a).

Note: In the question after finding a, b they can also verify them by putting a, b back in the expression. Then divide \[\left( x-1 \right)\] and \[\left( x-2 \right)\] individually to find remainder which are 0 and 2 respectively or not.