Question

If one of the zeroes of the quadratic polynomial \[(k - 1){x^2} + kx + 1\] is \[ - 3\], then \[k\] equals to:
A) \[\dfrac{4}{3}\]
B) \[ - \dfrac{4}{3}\]
C) \[\dfrac{2}{3}\]
D) \[ - \dfrac{2}{3}\]

Answer 1

Hint:
We will use the definition of the zeros of polynomials. We will substitute the given zero in the given polynomial and simplify it to get the value of \[k\]. A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using mathematical operations such as addition, subtraction, multiplication, and division.

Formula used:
If \[x\] is a zero of a polynomial, then \[p(x) = 0\].

Complete step by step solution:
The given polynomial is \[(k - 1){x^2} + kx + 1\]. This is a polynomial in \[x\].
Let \[p(x) = (k - 1){x^2} + kx + 1\] ……….\[(1)\]
Now, a zero of a polynomial is that value of \[x\] for which the polynomial vanishes i.e., \[p(x) = 0\].
It is given to us that \[ - 3\] is a zero of the polynomial \[p(x)\]. This means that when \[x = - 3\], we will get \[p( - 3) = 0\].
Let us put \[x = - 3\] in equation \[(1)\]. We get
\[p( - 3) = (k - 1){( - 3)^2} + k( - 3) + 1\]
Substituting \[p( - 3) = 0\] and simplifying the equation, we get
\[0 = 9(k - 1) - 3k + 1\]
Now multiplying the terms using the distributive property, we get
\[0 = 9k - 9 - 3k + 1\]
Adding like terms on the RHS, we get
\[\begin{array}{l}6k - 8 = 0\\ \Rightarrow 6k = 8\end{array}\]

Dividing both sides of the above equation by 6, we have
\[\begin{array}{l}\dfrac{{6k}}{6} = \dfrac{8}{6}\\ \Rightarrow k = \dfrac{4}{3}\end{array}\]

Therefore, the correct option is A.

Note:
The number of zeroes of a polynomial depends on the degree of the polynomial. The degree of a polynomial is the highest power of the variable in a polynomial equation. The polynomial given to us is quadratic. This means that the highest power of \[x\] in the polynomial is 2. Thus, there are at most 2 zeros of the polynomial.