Table of Contents
Exercise 3.2 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on zeros of polynomials. Mathematical expressions are rendered using MathJax.
1. The graphs of \( y = p(x) \) are given in the figure below, for some polynomials \( p(x) \). In each case, find the number of zeros of \( p(x) \).
The number of zeros of a polynomial corresponds to the number of times its graph intersects the x-axis (where \( y = 0 \)).
Since the graphs (i) to (vi) are not accessible, I cannot determine the exact number of intersections.
To solve this, you would typically count the number of x-axis intersections for each graph (i) through (vi).
Note: Please refer to the graphs in the textbook to count the number of zeros by observing the x-axis intersections.
2. Find the zeros of the given polynomials.
(i) \( p(x) = 3x \)
Set the polynomial equal to zero: \( 3x = 0 \).
Solve for \( x \): \( x = 0 \).
This is a linear polynomial (degree 1), so it has exactly 1 zero.
Zeros: 0
(ii) \( p(x) = x^2 + 5x + 6 \)
Set the polynomial equal to zero: \( x^2 + 5x + 6 = 0 \).
Factorize the quadratic: Find two numbers that multiply to 6 and add to 5.
The numbers are 2 and 3, so \( x^2 + 5x + 6 = (x + 2)(x + 3) \).
Set each factor to zero: \( x + 2 = 0 \implies x = -2 \), \( x + 3 = 0 \implies x = -3 \).
This is a quadratic polynomial (degree 2), so it has at most 2 zeros.
Zeros: -2, -3
(iii) \( p(x) = (x + 2)(x + 3) \)
The polynomial is already factored: \( (x + 2)(x + 3) = 0 \).
Set each factor to zero: \( x + 2 = 0 \implies x = -2 \), \( x + 3 = 0 \implies x = -3 \).
Expanding, \( p(x) = x^2 + 5x + 6 \), a quadratic polynomial with 2 zeros.
Zeros: -2, -3
(iv) \( p(x) = x^2 – 16 \)
Set the polynomial equal to zero: \( x^2 – 16 = 0 \).
This is a difference of squares: \( x^2 – 16 = (x – 4)(x + 4) \).
Set each factor to zero: \( x – 4 = 0 \implies x = 4 \), \( x + 4 = 0 \implies x = -4 \).
This is a quadratic polynomial (degree 2), so it has at most 2 zeros.
Zeros: 4, -4
3. Draw the graphs of the given polynomial and find the zeros. Justify the answers.
Since drawing graphs requires a visual tool, I will find the zeros algebraically and describe the graphing process.
To graph, you would typically: (1) Find the zeros, (2) Identify the y-intercept, (3) Plot additional points, (4) Sketch the curve based on the degree and leading coefficient.
(i) \( p(x) = x^2 – x – 12 \)
Set the polynomial equal to zero: \( x^2 – x – 12 = 0 \).
Factorize: Find two numbers that multiply to -12 and add to -1.
The numbers are -4 and 3, so \( x^2 – x – 12 = (x – 4)(x + 3) \).
Set each factor to zero: \( x – 4 = 0 \implies x = 4 \), \( x + 3 = 0 \implies x = -3 \).
For graphing: The parabola opens upward (leading coefficient is 1). Y-intercept is \( p(0) = -12 \). Vertex is at \( x = \frac{-b}{2a} = \frac{1}{2} \), \( p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 – \frac{1}{2} – 12 = -12.25 \).
The graph intersects the x-axis at \( x = -3 \) and \( x = 4 \), confirming the zeros.
Zeros: -3, 4
(ii) \( p(x) = x^2 – 6x + 9 \)
Set the polynomial equal to zero: \( x^2 – 6x + 9 = 0 \).
Factorize: \( x^2 – 6x + 9 = (x – 3)^2 \).
Set the factor to zero: \( (x – 3)^2 = 0 \implies x – 3 = 0 \implies x = 3 \).
This is a repeated root, so the zero is \( x = 3 \).
For graphing: The parabola opens upward. Y-intercept is \( p(0) = 9 \). Vertex is at \( x = \frac{6}{2} = 3 \), \( p(3) = 0 \), which is the zero.
The graph touches the x-axis at \( x = 3 \), confirming the repeated zero.
Zeros: 3 (repeated)
(iii) \( p(x) = x^2 – 4x + 5 \)
Set the polynomial equal to zero: \( x^2 – 4x + 5 = 0 \).
Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), \( c = 5 \).
Discriminant: \( b^2 – 4ac = (-4)^2 – 4 \cdot 1 \cdot 5 = 16 – 20 = -4 \).
Since the discriminant is negative, there are no real roots.
For graphing: The parabola opens upward. Y-intercept is \( p(0) = 5 \). Vertex is at \( x = \frac{4}{2} = 2 \), \( p(2) = 2^2 – 4 \cdot 2 + 5 = 1 \).
The graph does not intersect the x-axis, confirming no real zeros.
Zeros: None (no real zeros)
(iv) \( p(x) = x^2 + 3x – 4 \)
Set the polynomial equal to zero: \( x^2 + 3x – 4 = 0 \).
Factorize: Find two numbers that multiply to -4 and add to 3.
The numbers are 4 and -1, so \( x^2 + 3x – 4 = (x + 4)(x – 1) \).
Set each factor to zero: \( x + 4 = 0 \implies x = -4 \), \( x – 1 = 0 \implies x = 1 \).
For graphing: The parabola opens upward. Y-intercept is \( p(0) = -4 \). Vertex is at \( x = \frac{-3}{2} = -1.5 \), \( p(-1.5) = (-1.5)^2 + 3(-1.5) – 4 = -6.25 \).
The graph intersects the x-axis at \( x = -4 \) and \( x = 1 \), confirming the zeros.
Zeros: -4, 1
(v) \( p(x) = x^2 – 1 \)
Set the polynomial equal to zero: \( x^2 – 1 = 0 \).
Factorize: \( x^2 – 1 = (x – 1)(x + 1) \).
Set each factor to zero: \( x – 1 = 0 \implies x = 1 \), \( x + 1 = 0 \implies x = -1 \).
For graphing: The parabola opens upward. Y-intercept is \( p(0) = -1 \). Vertex is at \( x = 0 \), \( p(0) = -1 \).
The graph intersects the x-axis at \( x = -1 \) and \( x = 1 \), confirming the zeros.
Zeros: -1, 1
4. Why are \( \frac{1}{4} \) and -1 zeros of the polynomial \( p(x) = 4x^2 + 3x – 1 \)?
To confirm \( \frac{1}{4} \) and -1 are zeros, substitute them into the polynomial and check if \( p(x) = 0 \).
For \( x = \frac{1}{4} \): \( p\left(\frac{1}{4}\right) = 4\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) – 1 = 4 \cdot \frac{1}{16} + \frac{3}{4} – 1 = \frac{4}{16} + \frac{3}{4} – 1 = \frac{1}{4} + \frac{3}{4} – 1 = 1 – 1 = 0 \).
For \( x = -1 \): \( p(-1) = 4(-1)^2 + 3(-1) – 1 = 4 \cdot 1 – 3 – 1 = 4 – 3 – 1 = 0 \).
Since \( p\left(\frac{1}{4}\right) = 0 \) and \( p(-1) = 0 \), they are zeros.
Alternatively, factorize: \( 4x^2 + 3x – 1 = 0 \). Using the quadratic formula or trial, factor as \( (4x – 1)(x + 1) \).
Check: \( (4x – 1)(x + 1) = 4x^2 + 4x – x – 1 = 4x^2 + 3x – 1 \), which matches.
Solve: \( 4x – 1 = 0 \implies x = \frac{1}{4} \), \( x + 1 = 0 \implies x = -1 \).
Reason: \( \frac{1}{4} \) and -1 satisfy \( p(x) = 0 \), as shown by substitution and factorization.
