Question

Verify whether the following are zeroes of the polynomial, indicated against them.
(i)$p(x) = 3x + 1,x = - \dfrac{1}{3}$
(ii)$p(x) = 5x - \pi ,x = \dfrac{4}{5}$
(iii)$p(x) = {x^2} - 1,x = 1, - 1$
(iv)$p(x) = (x + 1)(x - 2),x = - 1,2$
(v)$p(x) = {x^2},x = 0$
(vi)$p(x) = lx + m,x = - \dfrac{m}{l}$
(vii)$p(x) = 3{x^2} - 1,x = - \dfrac{1}{{\sqrt 3 }},\dfrac{2}{{\sqrt 3 }}$
(viii)$p(x) = 2x + 1,x = \dfrac{1}{2}$

Answer 1

Hint: The zeroes of a polynomial refer to the numerical values of the variable corresponding to which the value of the polynomial expression is zero. For example if $a$ is a zero of a polynomial $p(x)$ ,then $p(a) = 0$ . To find the zeros of a polynomial, set the polynomial expression equal to zero and then solve that equation for the given variable.

Complete step-by-step answer:
For the questions we will be following the process mentioned in the Solution Hint.
(i):
 We need to find the value of $x$ for which $p(x)$ is $0$.
So setting $p(x) = 0$ , we get,
$3x + 1 = 0$
$ \Rightarrow x = - \dfrac{1}{3}$
Therefore, $x = - \dfrac{1}{3}$ is a zero of polynomials $p(x)$.
(ii):
Again, we set $p(x) = 0$ ,
$ \Rightarrow 5x - \pi = 0$
$ \Rightarrow x = \dfrac{\pi }{5}$
Therefore, the zero of polynomial $5x - \pi $ is $x = \dfrac{\pi }{5}$ and not $x = \dfrac{4}{5}$.
(iii):
Like before, we set $p(x) = 0$,
${x^2} - 1 = 0$
$ \Rightarrow {x^2} = 1$
$ \Rightarrow x = \sqrt 1 $
$ \Rightarrow x = \pm 1$ (Because both $ - 1 \times - 1 = 1$ and $1 \times 1 = 1$ )
So the zeroes of this polynomial are indeed $x = + 1$ and $x = - 1$. Note that this has two zeroes because it is a quadratic polynomial.
(iv):
We set $p(x) = 0$ like before,
$ \Rightarrow (x + 1)(x - 2) = 0$
Now since it is a multiplication of two terms, the product will be zero when either of the terms are zero. So we will set both the terms to zero individually and get the two zeroes of this polynomial.
Setting $x + 1 = 0$ , we get, $x = - 1$.
And, setting, $x - 2 = 0$, we get, $x = 2$.
Therefore the two zeroes of this polynomial are $x = - 1$ and $x = 2$.
(v):
Setting $p(x) = 0$ ,we get,
${x^2} = 0$
$ \Rightarrow x = 0$ ($\because \sqrt 0 = 0$ )
So the zero of this polynomial is $x = 0$.
(vi):
In this question, $l$ and $m$ are coefficients. Like the previous questions, we set $p(x) = 0$.
$\therefore lx + m = 0$
$ \Rightarrow lx = - m$
$ \Rightarrow x = - \dfrac{m}{l}$
Therefore, the zero of this polynomial is $x = - \dfrac{m}{l}$.
(vii):
Setting $p(x) = 0$ ,we get,
$3{x^2} - 1 = 0$
$ \Rightarrow 3{x^2} = 1$
$ \Rightarrow {x^2} = \dfrac{1}{3}$
$ \Rightarrow x = \dfrac{{\sqrt 1 }}{{\sqrt 3 }}$
$ \Rightarrow x = \pm \dfrac{1}{{\sqrt 3 }}$
The zeroes of this polynomial are, therefore, $x = \dfrac{1}{{\sqrt 3 }}$ and $x = - \dfrac{1}{{\sqrt 3 }}$ and not $x = \dfrac{2}{{\sqrt 3 }}$.
(viii):
Setting $p(x) = 0$ , we get,
$2x + 1 = 0$
$ \Rightarrow 2x = - 1$
$ \Rightarrow x = - \dfrac{1}{2}$
Therefore, the zero of this polynomial is $x = - \dfrac{1}{2}$ and not $x = \dfrac{1}{2}$.

Note: Always remember that when taking the square root of a number, we need to consider both the positive and the negative component of it. Also observe that for a linear polynomial we get one zero, for a quadratic polynomial, we get two zeroes, for a cubic polynomial, we get three zeroes and so on. So if you get more or less a number of zeroes for a polynomial of a particular degree, recheck your answer for possible mistakes.