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Q3(iii):
Verify whether the following are zeroes of the polynomial, indicated against them.
(iii) $p(x) = x^2 – 1, x = 1, –1$
Solution :
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 $p(x) = x^2 - 1$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2
- Q1(i): Find the value of the polynomial $5x – 4x^2 + 3$ at (i) $x = 0$
- Q1(ii): Find the value of the polynomial $5x – 4x^2 + 3$ at (ii) $x = –1$
- Q1(iii): Find the value of the polynomial $5x – 4x^2 + 3$ at (iii) $x = 2$
- Q2(i): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (i) $p(y) = y^2 – y + 1$
- Q2(ii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (ii) $p(t) = 2 + t + 2t^2 – t^3$
- Q2(iii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iii) $p(x) = x^3$
- Q2(iv): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iv) $p(x) = (x – 1) (x + 1)$
- Q3(i): Verify whether the following are zeroes of the polynomial, indicated against them. (i) $p(x) = 3x + 1, x = –\frac{1}{3}$
- Q3(ii): Verify whether the following are zeroes of the polynomial, indicated against them. (ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
- Q3(iv): Verify whether the following are zeroes of the polynomial, indicated against them. (iv) $p(x) = (x + 1) (x – 2), x = – 1, 2$
- Q3(v): Verify whether the following are zeroes of the polynomial, indicated against them. (v) $p(x) = x^2, x = 0$
- Q3(vi): Verify whether the following are zeroes of the polynomial, indicated against them. (vi) $p(x) = lx + m, x = –\frac{m}{l}$
- Q3(vii): Verify whether the following are zeroes of the polynomial, indicated against them. (vii) $p(x) = 3x^2 – 1, x = –\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
- Q3(viii): Verify whether the following are zeroes of the polynomial, indicated against them. (viii) $p(x) = 2x + 1, x = \frac{1}{2}$
- Q4(i): Find the zero of the polynomial in each of the following cases: (i) $p(x) = x + 5$
- Q4(ii): Find the zero of the polynomial in each of the following cases: (ii) $p(x) = x – 5$
- Q4(iii): Find the zero of the polynomial in each of the following cases: (iii) $p(x) = 2x + 5$
- Q4(iv): Find the zero of the polynomial in each of the following cases: (iv) $p(x) = 3x – 2$
- Q4(v): Find the zero of the polynomial in each of the following cases: (v) $p(x) = 3x$
- Q4(vi): Find the zero of the polynomial in each of the following cases: (vi) $p(x) = ax, a \neq 0$
- Q4(vii): Find the zero of the polynomial in each of the following cases: (vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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