Q3(iii):

Verify whether the following are zeroes of the polynomial, indicated against them. (iii) $p(x) = x^2 – 1, x = 1, –1$

Initial Setup & Theoretical Foundation

We are given the quadratic polynomial:

$p(x) = x^2 - 1$

We must verify whether the given values, $x = 1$ and $x = -1$, are zeroes of the polynomial $p(x)$.

Theoretical Justification: [Per the Definition of a Zero of a Polynomial], a real number $c$ is considered a zero (or root) of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = c$ yields zero. Mathematically, this is expressed as $p(c) = 0$. Furthermore, [By the Factor Theorem], if $p(c) = 0$, then $(x - c)$ is a factor of the polynomial.

Step 1: Evaluating the Polynomial at $x = 1$

To determine if $x = 1$ is a zero, we substitute $x = 1$ into the polynomial $p(x)$:

• $p(1) = (1)^2 - 1$
• $p(1) = 1 - 1$
• $p(1) = 0$

Conclusion for $x = 1$: Since the evaluation results in exactly $0$, $x = 1$ is a verified zero of the polynomial $p(x)$.

Step 2: Evaluating the Polynomial at $x = -1$

Next, we substitute $x = -1$ into the polynomial $p(x)$:

• $p(-1) = (-1)^2 - 1$
• $p(-1) = 1 - 1$ [Since the square of any negative real number is positive, $(-1) \times (-1) = 1$]
• $p(-1) = 0$

Conclusion for $x = -1$: Since the evaluation results in exactly $0$, $x = -1$ is also a verified zero of the polynomial $p(x)$.

Graphical Verification (Visualizing the Zeroes)

In coordinate geometry, the real zeroes of a polynomial $p(x)$ correspond precisely to the $x$-intercepts of the graph of the equation $y = p(x)$. By plotting the parabola $y = x^2 - 1$, we can visually confirm that the curve intersects the $x$-axis at exactly $x = -1$ and $x = 1$.

Alternative Algebraic Perspective (Factorization)

We can also verify the zeroes by factoring the polynomial completely. The expression $x^2 - 1$ is a classic "Difference of Squares", which follows the algebraic identity $a^2 - b^2 = (a - b)(a + b)$.

$p(x) = x^2 - 1^2$

$p(x) = (x - 1)(x + 1)$

To find the zeroes, we set $p(x) = 0$:

$(x - 1)(x + 1) = 0$

[By the Zero Product Property], if the product of two factors is zero, at least one of the factors must be zero:

• $x - 1 = 0 \implies x = 1$
• $x + 1 = 0 \implies x = -1$

This algebraic derivation perfectly matches our initial substitution method.

Final Solution: Yes, both $x = 1$ and $x = -1$ are verified zeroes of the polynomial $p(x) = x^2 - 1$.