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Q2(iv):
Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials:
(iv) $p(x) = (x – 1) (x + 1)$
Solution :
Initial Setup & Polynomial Analysis
We are given the polynomial function defined as:
$p(x) = (x - 1)(x + 1)$
Before evaluating the polynomial at specific discrete points, we can analyze its structure. The expression is factored into two binomials. [Per the algebraic identity for the difference of two squares: $(a - b)(a + b) = a^2 - b^2$], we can expand the polynomial into its standard quadratic form:
$p(x) = x^2 - 1^2 = x^2 - 1$
This represents a parabola opening upwards with its vertex at $(0, -1)$. We will evaluate the polynomial using both the factored form and the expanded form to ensure absolute mathematical rigor.
Step 1: Evaluating $p(0)$
To find the value of the polynomial when the independent variable $x$ is exactly $0$, we substitute $0$ into the function.
- Using the factored form:
$p(0) = (0 - 1)(0 + 1)$
$p(0) = (-1)(1)$
$p(0) = -1$ - Using the expanded form [Verification]:
$p(0) = (0)^2 - 1 = 0 - 1 = -1$
Geometrically, this represents the y-intercept of the polynomial curve at the coordinate $(0, -1)$.
Step 2: Evaluating $p(1)$
Next, we substitute $x = 1$ into the polynomial.
- Using the factored form:
$p(1) = (1 - 1)(1 + 1)$
$p(1) = (0)(2)$
$p(1) = 0$ - Using the expanded form [Verification]:
$p(1) = (1)^2 - 1 = 1 - 1 = 0$
Because $p(1) = 0$, $x = 1$ is a root (or zero) of the polynomial. [By the Factor Theorem, since $(x - 1)$ is a factor of $p(x)$, $p(1)$ must equal zero].
Step 3: Evaluating $p(2)$
Finally, we substitute $x = 2$ into the polynomial.
- Using the factored form:
$p(2) = (2 - 1)(2 + 1)$
$p(2) = (1)(3)$
$p(2) = 3$ - Using the expanded form [Verification]:
$p(2) = (2)^2 - 1 = 4 - 1 = 3$
This gives us the coordinate point $(2, 3)$ on the Cartesian plane.
Graphical Representation of the Polynomial
Below is the precise geometric plot of the quadratic function $p(x) = x^2 - 1$. The specific points we evaluated—$p(0)$, $p(1)$, and $p(2)$—are explicitly mapped to demonstrate the relationship between algebraic evaluation and spatial coordinates.
Final Solution: The evaluated values for the polynomial $p(x) = (x - 1)(x + 1)$ are $p(0) = -1$, $p(1) = 0$, and $p(2) = 3$.
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) $p(x) = x^3$
- 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(iii): Verify whether the following are zeroes of the polynomial, indicated against them. (iii) $p(x) = x^2 – 1, x = 1, –1$
- 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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