Question

If \[ - 1\] is a zero of the polynomial \[p\left( x \right) = a{x^3} - {x^2} + x + 4\], then the value of \[a\].

Answer 1

Hint:
 The given polynomial is a cubic polynomial. Substitute the value \[x = - 1\] in the polynomial \[p\left( x \right) = a{x^3} - {x^2} + x + 4\] and equate it to 0. Simplify the equation and solve for the value of \[a\].

Complete step by step solution:
The given polynomial is a cubic polynomial.
The zero of a polynomial is that number which when substituted in the polynomial, gives the value 0.
If \[ - 1\] is a zero, then after substituting \[x = - 1\] in the polynomial \[p\left( x \right) = a{x^3} - {x^2} + x + 4\] we will get 0 as the answer.
Then, \[p\left( { - 1} \right) = a{\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} + \left( { - 1} \right) + 4\]
Also, \[p\left( { - 1} \right) = 0\]
\[a{\left( { - 1} \right)^3} - {\left( { - 1} \right)^2} + \left( { - 1} \right) + 4 = 0\]
Solve the brackets.
$
   - a - 1 - 1 + 4 = 0 \\
   - a + 2 = 0 \\
$
On taking 2 on the other side it will become \[ - 2\]. Then divide the equation throughout by \[ - 2\].
$
   - a = - 2 \\
  a = 2 \\
$

Hence, the value of \[a\] is 2.

Note:
The value of the variables for the polynomial results to 0 are known as the zeroes of the polynomial. A polynomial with degree \[n\] has at most \[n\] zeroes. All the linear polynomials can have only one zero. In this question, we have a cubic polynomial.