# 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 $a$.

Last updated date: 17th Mar 2023

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Answer

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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 $a$.

Complete step-by-step answer:

We have been given that $p\left( x \right) = a{x^2} - 3\left( {a - 1} \right)x - 1$……….. (1)

Now putting $x = 1$in equation (1) we get,

$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1$ ………… (2)

Since 1 is zero of the polynomial $p\left( x \right)$ hence at $x = 1$the value of $p\left( 1 \right) = 0$

From equation (2) we have,

$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1 = 0$……….. (3)

Now simplify the equation (3) we get,

$

\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 $a = 1$.

Note: Whenever we face such types of problems the key concept is by putting the value of variable in the polynomial equal to the given zeros of the polynomial and equate the polynomial to zero at the given zeros to get the value of constant used in the polynomial.

Complete step-by-step answer:

We have been given that $p\left( x \right) = a{x^2} - 3\left( {a - 1} \right)x - 1$……….. (1)

Now putting $x = 1$in equation (1) we get,

$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1$ ………… (2)

Since 1 is zero of the polynomial $p\left( x \right)$ hence at $x = 1$the value of $p\left( 1 \right) = 0$

From equation (2) we have,

$p\left( 1 \right) = a{\left( 1 \right)^2} - 3\left( {a - 1} \right)\left( 1 \right) - 1 = 0$……….. (3)

Now simplify the equation (3) we get,

$

\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 $a = 1$.

Note: Whenever we face such types of problems the key concept is by putting the value of variable in the polynomial equal to the given zeros of the polynomial and equate the polynomial to zero at the given zeros to get the value of constant used in the polynomial.

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