Question

If $f\left( x \right)={{x}^{3}}+{{x}^{2}}-ax+b$ is divisible by ${{x}^{2}}-x$, write the value of a and b.

Answer 1

Hint: We can see that the expression ${{x}^{2}}-x$ can be factored as $x\left( x-1 \right)$. If it is given that one number is divided by another, then we can write the dividend as the product of quotient and divisor plus remainder. But since it completely divides it is given that ${{x}^{2}}-x$ divides f(x), this means the remainder is 0. Thus, the product of ${{x}^{2}}-x$ and one another factor of f(x) is equal to f(x). Then, to find the zeros, we will put f(x) = 0. This will give us values of x for which f(x) is zero. We will then find two such values out of three and put them in f(x) so that we can form two equations with a and b. Once we get those equations, we can solve them to find the value of a and b.

Complete step by step answer:
It is given to us that $f\left( x \right)={{x}^{3}}+{{x}^{2}}-ax+b$ and it is divisible by ${{x}^{2}}-x$.
This means, the product of ${{x}^{2}}-x$ and some other factor of f(x) is equal to f(x).
Let that other factor be g(x).
$\Rightarrow \left( {{x}^{2}}-x \right)g\left( x \right)=f\left( x \right)$
Now, we know that ${{x}^{2}}-x$ factorises as $x\left( x-1 \right)$.
$\Rightarrow x\left( x-1 \right)g\left( x \right)=f\left( x \right)$
To find the zero of f(x), we will put f(x) = 0.
$\Rightarrow x\left( x-1 \right)g\left( x \right)=0$
This means, either x = 0 or x – 1 = 0 or g(x) = 0.
Therefore, we are sure that if we put x = 0 or x = 1, the value of f(x) = 0.
Thus, substitute x = 0 in the f(x).
$\begin{align}
  & \Rightarrow f\left( 0 \right)={{\left( 0 \right)}^{3}}+{{\left( 0 \right)}^{2}}-a\left( 0 \right)+b \\
 & \Rightarrow b=0 \\
\end{align}$
Now, we substitute x = 1.
$\Rightarrow f\left( 1 \right)={{\left( 1 \right)}^{3}}+{{\left( 1 \right)}^{2}}-a\left( 1 \right)+b$
But we have shown that b = 0
$\begin{align}
  & \Rightarrow 0=1+1-a+0 \\
 & \Rightarrow a=2 \\
\end{align}$

Therefore, a = 2 and b = 0.

Note: The number of zeros of a polynomial is equal to the highest power of that polynomial. Thus, a quadratic expression has two zeros, a cubic expression has three zeros and so on.