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Q2(ii):
Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials:
(ii) $p(t) = 2 + t + 2t^2 – t^3$
Solution :
Given Variables & Initial Setup
We are given the following cubic polynomial in terms of the variable $t$:
$p(t) = 2 + t + 2t^2 - t^3$
To find the values of $p(0)$, $p(1)$, and $p(2)$, we must apply the Polynomial Evaluation Principle. This principle states that the value of a polynomial $p(t)$ at a specific real number $t = a$ is obtained by substituting $a$ for every instance of $t$ in the polynomial expression [Per the fundamental definition of a polynomial function].
Step 1: Evaluating $p(0)$
We substitute $t = 0$ into the polynomial equation. This operation isolates the constant term, as all terms containing the variable $t$ will evaluate to zero.
- $p(0) = 2 + (0) + 2(0)^2 - (0)^3$
- $p(0) = 2 + 0 + 2(0) - 0$
- $p(0) = 2$
Geometrically, this represents the $y$-intercept of the polynomial graph at the coordinate $(0, 2)$.
Step 2: Evaluating $p(1)$
Next, we substitute $t = 1$ into the polynomial. Evaluating a polynomial at $t = 1$ effectively yields the sum of its coefficients and the constant term.
- $p(1) = 2 + (1) + 2(1)^2 - (1)^3$
- $p(1) = 2 + 1 + 2(1) - 1$
- $p(1) = 3 + 2 - 1$
- $p(1) = 4$
This corresponds to the coordinate point $(1, 4)$ on the Cartesian plane.
Step 3: Evaluating $p(2)$
Finally, we substitute $t = 2$ into the polynomial. We must carefully follow the order of operations (PEMDAS/BODMAS), evaluating the exponents before multiplying by the coefficients.
- $p(2) = 2 + (2) + 2(2)^2 - (2)^3$
- $p(2) = 2 + 2 + 2(4) - 8$
- $p(2) = 4 + 8 - 8$
- $p(2) = 4$
This corresponds to the coordinate point $(2, 4)$ on the Cartesian plane.
Graphical Representation of $p(t)$
Below is the precise Cartesian plot of the polynomial $p(t) = -t^3 + 2t^2 + t + 2$. The specific points we evaluated—$p(0)$, $p(1)$, and $p(2)$—are explicitly marked to demonstrate their spatial relationship on the curve.
Summary of Results
| Variable Input ($t$) | Polynomial Output ($p(t)$) |
|---|---|
| $t = 0$ | $2$ |
| $t = 1$ | $4$ |
| $t = 2$ | $4$ |
Final Solution: The evaluated values for the polynomial are $p(0) = 2$, $p(1) = 4$, and $p(2) = 4$.
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(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(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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