Question

Find the value of the polynomial $5x-4{{x}^{2}}+3$ at $x=0$.
(a) 3
(b) 2
(c) 0
(d) 1

Answer 1

Hint: The value of the polynomial can easily be obtained using the given condition. In cases where the condition is given, the best and easiest way to solve is replacing the variables in the polynomial expression with the values mentioned in the condition where the value of the polynomial is to be found out. Here, the given condition is given as $x=0$, hence, we will have to replace the value of x with ‘zero’ in the given polynomial equation as it is the given condition where we have to find the value of the polynomial expression $5x-4{{x}^{2}}+3=0$.

Complete step-by-step solution:
The given polynomial expression is $5x-4{{x}^{2}}+3=0.........................(i)$
We have to find the value of the polynomial expression mentioned in equation (i) at $x=0$. Now, we can find the value of the expression by replacing the value of x in the polynomial equation given above.
Let us firstly assume $P(x)=5x-4{{x}^{2}}+3...................(ii)$
Now, we replace the value of x with ‘zero’ in the above-given equation (ii), then, we get,
$P\left( 0 \right)=5\times 0-4\times {{\left( 0 \right)}^{2}}+3$
$P\left( 0 \right)=0-0+3$
$P\left( 0 \right)=3$
Hence, on replacing the value of x with ‘zero’, we get the value of the given expression as 3, which is the required answer.
Hence, the correct option is option (a).

Note: Students mostly make mistakes will replacing the value and doing the calculation.
An alternate method, to solve this question is to use the remainder theorem. The remainder theorem implies that divide the equation (i) with x. And the remainder left after the entire division process will give the value of the polynomial.
Dividing the polynomial expression $5x-4{{x}^{2}}+3$ with x, we get,
$\Rightarrow x\overline{\left){5x-4{{x}^{2}}+3}\right.}$
$\Rightarrow x\overset{5}{\overline{\left){\begin{align}
  & 5x-4{{x}^{2}}+3 \\
 & \underline{-5x\text{ }} \\
 & 0-4{{x}^{2}} \\
\end{align}}\right.}}$
$\Rightarrow x\overset{5+4x}{\overline{\left){\begin{align}
  & 5x-4{{x}^{2}}+3 \\
 & \underline{-5x\text{ }} \\
 & 0-4{{x}^{2}} \\
 & \underline{\text{ +4}{{\text{x}}^{2}}\text{ }}\text{ } \\
 & \text{ 0 } \\
\end{align}}\right.}}$
$\Rightarrow x\overset{5+4x}{\overline{\left){\begin{align}
  & 5x-4{{x}^{2}}+3 \\
 & \underline{-5x\text{ }} \\
 & 0-4{{x}^{2}} \\
 & \underline{\text{ +4}{{\text{x}}^{2}}\text{ }}\text{ } \\
 & \text{ }0+3\text{ } \\
\end{align}}\right.}}$
$\Rightarrow x\overset{5+4x}{\overline{\left){\begin{align}
  & 5x-4{{x}^{2}}+3 \\
 & \underline{-5x\text{ }} \\
 & 0-4{{x}^{2}} \\
 & \underline{\text{ +4}{{\text{x}}^{2}}\text{ }}\text{ } \\
 & \text{ +}3\text{ } \\
\end{align}}\right.}}$
Since the remainder is 3, so, the value of the expression is 3 at $x=0$.