Madhya Pradesh Police Exam Model Paper 5

Tips & Tricks
TB Solution-
Concept Used- Factorisation of a quadratic equation.
Given Expression $={\mathit{x}}^2-9\mathit{x}+20$
Calculation-
$={\mathit{x}}^2-9\mathit{x}+20$
$={\mathit{x}}^2-5\mathit{x}-4\mathit{x}+20$
$=\mathit{x}\left(\mathit{x}-5\right)-4\left(\mathit{x}-5\right)$
$=\left(\mathit{x}-4\right)\left(\mathit{x}-5\right)$
∴ From the given options $\left(\mathit{x}-4\right)$ is factor of ${\mathit{x}}^2-9\mathit{x}+20$
Short trick-
From the given options, we can put the different values of x in ${\mathit{x}}^2-9\mathit{x}+20$ and check for which value ${\mathit{x}}^2-9\mathit{x}+20=0$
Putting $\mathit{x}=4$ , we get ${\mathit{x}}^2-9\mathit{x}+20=0$
∴ $\mathit{x}-4$ is a factor of ${\mathit{x}}^2-9\mathit{x}+20$.