Question

# If 1 is zero of the polynomial $p\left( x \right) = a{x^2} - 3\left( {a - 1} \right)x - 1$, then find the value of $a$.

Hint: In this question, it is given that 1 is the zero of polynomial $p\left( x \right) = a{x^2} - 3\left( {a - 1} \right)x - 1$ which means that if we put $x = 1$in polynomial $p\left( x \right)$ then we have $p\left( 1 \right) = 0$ and then we get the value of $a$.

We have been given that $p\left( x \right) = a{x^2} - 3\left( {a - 1} \right)x - 1$â€¦â€¦â€¦.. (1)
Now putting $x = 1$in equation (1) we get,
$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1$ â€¦â€¦â€¦â€¦ (2)
Since 1 is zero of the polynomial $p\left( x \right)$ hence at $x = 1$the value of $p\left( 1 \right) = 0$
$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1 = 0$â€¦â€¦â€¦.. (3)
$\Rightarrow a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1 = 0 \\ \Rightarrow a - 3\left( {a - 1} \right) - 1 = 0 \\ \Rightarrow a - 3a + 3 - 1 = 0 \\ \Rightarrow - 2a + 2 = 0 \\ \Rightarrow - 2a = - 2 \\ \Rightarrow a = \dfrac{{ - 2}}{{ - 2}} \\ \Rightarrow a = 1 \\$
And hence the value of $a = 1$.