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Q1(i):
Find the value of the polynomial $5x – 4x^2 + 3$ at
(i) $x = 0$
Solution :
Initial Setup & Polynomial Definition
Let the given mathematical expression be defined as a polynomial function $P(x)$.
$P(x) = 5x - 4x^2 + 3$
[Per the standard form of a polynomial, it is conventionally written in descending order of the degree of its terms. Rearranging the terms yields:]
$P(x) = -4x^2 + 5x + 3$
Step 1: Substitution of the Variable
We are required to evaluate the polynomial at the specific coordinate $x = 0$. This process involves substituting the value $0$ for every instance of the variable $x$ in the polynomial expression.
$P(0) = -4(0)^2 + 5(0) + 3$
Step 2: Algebraic Evaluation
Apply the standard order of operations [PEMDAS/BODMAS: evaluating exponents first, followed by multiplication, and finally addition].
| Polynomial Term | Substitution ($x = 0$) | Evaluated Value |
|---|---|---|
| $-4x^2$ | $-4(0)^2 = -4(0)$ | $0$ |
| $5x$ | $5(0)$ | $0$ |
| $+3$ (Constant Term) | $+3$ | $3$ |
Summing the evaluated terms:
$P(0) = 0 + 0 + 3$
$P(0) = 3$
Graphical Interpretation (The Y-Intercept)
Geometrically, evaluating any polynomial $P(x)$ at $x = 0$ yields the y-intercept of its graph. The constant term of the polynomial represents the exact point where the curve crosses the y-axis. For the quadratic function $P(x) = -4x^2 + 5x + 3$, the graph is a downward-opening parabola that intersects the y-axis at the coordinate $(0, 3)$.
Final Solution: The value of the polynomial $5x - 4x^2 + 3$ at $x = 0$ is 3.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2
- 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(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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