Telangana SCERT Class 10 Maths Chapter 3 Polynomials Exercise 3.2 Maths Problems and Solution Here in this Post. Telangana SCERT Class 10 Maths Solution Chapter 3 Polynomials Exercise 3.2
Telangana SCERT Solution Class X (10) Maths Chapter 3 Polynomials Exercise 3.2
Exercise 3.2
(Q1) The graphs of y = p(x) are given in the figures below, for some polynomials p(x). In each case, find the number of zeroes of p(x).
(i)
Number of zeroes = 0
We have to observe whether the curve is interesting the x – axis or not. If it is intersect the x – axis then how many points it is interesting the x – axis now have to count so we can see the graph there is no intersect the x – axis.
Hence, number of zeroes = 0
(ii)
Number of zeroes = 1
We can see the graph, there is only one point is intersect on x – axis.
So, number of zeroes = 1
(iii)
We can the graph, there are three distinct points intersecting the graph.
So, number of zeroes = 3
[The curve is intersecting on x – axis or not. If curve is intersecting on x – axis the curve how many points the curve is intersect. The curve is intersect only x – axis number of zeroes depend on x – axis.
(iv)
We can see the graph is intersecting in two distinct points on x – axis.
So number of zeroes = 2
[Count the number of points of intersection]
(v)
We can see the graph, the curve is intersecting the four distinct points on x – axis.
∴ Number of zeroes = 4
(vi)
We can see the graph, the curve is intersecting the three distinct points on x – axis.
So, number of zeroes = 3
(Q2) Find the zeroes of the given polynomials.
(i) p(x) = 3x
=> solution:
Let p(x) = 0
=> 3x = 0 (given)
=> x = 0/3
=> x = 0
∴ Zero of p(x) = 0
(ii) p(x) = x2 + 5x + 6
=> Solution:
Let p(x) = 0
=> x2 + 5x + 6 = 0 (Given)
To find the factor of 6
+6
+3 +2
=> x2 + 3x + 2x + 6 = 0
=> x (x + 3) +2 (x + 3) = 0
=> (x + 3) (x + 2) = 0
=> x + 3 = 0 or x + 2 = 0
=> x = – 3 or x = – 2
∴ Zeroes of p(x) = – 3, – 2
(iii) P(x) = (x + 2) (x + 3)
=> Solution:
Let p (x) = 0
=> (x + 2) (x + 3) = 0 (Given)
=> x + 2 = 0 or x + 3 = 0
=> x = – 2 or x = – 3
∴ Zeroes of p(x) = – 2, – 3
(iv) p(x) = x4 – 16
=> Solution:
Let p(x) = 0
=> x4 – 16 = 0 (Given)
=> (x2)2 – 42 = 0
We can write a2 + b2 form
So, a2 + b2 = (a + b) (a – b)
=> (x2 + 4) (x2 – 4) = 0
=> (x2 + 4) (x2 – 22) = 0
=> (x2 + 4) (x + 2) (x – 2) = 0
=> x2 + 4 = 0 or x + 2 = 0 or x – 2 = 0
=> x2 = – 4, x = – 2, x = 2
=> x = ± √-4, – 2, 2
∴ Zeroes of p(x) = ± √-4, – 2, 2
(Q3) Draw the graphs of the given polynomial and find the zeroes. Justify the answers.
(i) p(x) = x2 – x – 12
=> Solution: The given polynomial.
P(x) = x2 – x – 12
Let y = x2 – x – 12
* Table of ordered pairs
x | -3 | -2 | 0 | 1 | 2 | 3 | 4 |
X2 | 9 | 4 | 0 | 1 | 4 | 9 | 16 |
X2 – x | 9 – (-3) =12 | 4 – (-2) =12 | 0 | 1 – 1 =0 | 4-2 =2 | 9-3 =6 | 16 – 4 = 12 |
Y = x2 – x – 12 | 12 – 12 =0 | 6-12 = -6 | 0-12 = -12 | 0 – 12 = -12 | 2 – 12 =-10 | 6-12 =-6 | 12 – 12 = 0 |
(x, y) | (-3, 0) | (-2, -6) | (0, -12) | (1, -12) | (2, -10) | (3, -6) | (4, 0) |
[Substitute x= -3, -2, 0, 1, 2, 3, 4 in given terms of polynomial]
By plotting points on graph,
(x,y) = (-3, 0), (-2,-6), (0,-12), (1,-12), (2,-10) (3,-6), (4,0)
(ii) p(x) = x2 – 6x +9
=> Solution: The given polynomial, p(x) = x2 – 6x +9
Let y = x2 – 6x + 9
* The table of ordered pairs
x | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
X2 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
6x | 6X-2 = -12 | 6X1 = 6 | 0 | 6 | 6X2 = 12 | 6X3 = 18 | 6X4 = 24 |
X2 – 6x | 16 | 7 | 0 | -5 | -8 | -9 | -8 |
x2 – 6x + 9 = y | 25 | 16 | 9 | 4 | 1 | 0 | 1 |
(x, y) | (-2,25) | (-1,16) | (0, 9) | (1,4) | (2,1) | (3,0) | (4,1) |
By plotting the points (-2,25), (-1,16), (0,9), (1,4), (2,1), (3,0), (4,1) on the graph.
(iii) P(x) = x2 – 4x + 5
=> Solution: The given polynomial,
P(x) = x2 – 4x + 5
Let y = x2 – 4x + 5
* Table of ordered pairs.
x | -3 | -1 | 0 | 1 | 2 | 3 | 4 |
x2 | 9 | 1 | 0 | 1 | 4 | 9 | 16 |
4x | -12 | -4 | 0 | 4 | 8 | 12 | 16 |
X2 – 4x | 21 | 5 | 0 | -3 | -4 | -3 | 0 |
Y= x2 – 4x= +5 | 26 | 10 | 5 | 2 | 1 | 2 | 5 |
(x, y) | (-3,26) | (-1,10) | (0,5) | (1,2) | (2,1) | (3,2) | (4,5) |
By plotting the pairs (-3,26), (-1,10), (0,5), (1,2), (2,1) (3,2), (4,5) on graph.
(iv) p(x) = x2 + 3x – 4
=> Solution: Give polynomial is,
P(x) = x2 + 3x – 4
Let y = x2 + 3x – 4
* Table of ordered pairs:
x | -4 | -3 | -2 | 0 | 1 | 2 | 3 |
x2 | 16 | 9 | 4 | 0 | 1 | 4 | 9 |
3x | -12 | -9 | -6 | 0 | 3 | 6 | 9 |
X2 + 3x | 4 | 0 | -2 | 0 | 4 | 10 | 18 |
Y = x2 + 3x – 4 | 0 | -4 | -6 | -4 | 0 | 6 | 14 |
(x, y) | (4,10) | (-3,4) | (-2,-6) | (0,-4) | (1,10) | (2,6) | (3,14) |
By plotting the points (-4,0), (-3,-4), (-2,-6), (0,-4), (1,0), (2,6), (3,14) on the graph.
(v) P (x) = x2 – 1
=> Solution: Given polynomial is,
P(x) = x2 – 1
Let y = x2 – 1
* Table of ordered pairs:
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
x2 | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
x2 – 1 =y | 8 | 3 | 0 | -1 | 0 | 3 | 8 |
(X, y) | (-3,8) | (-2,3) | =(-1,0) | (0,-1) | (1,0) | (2,3) | (3,8) |
By plotting the points (-3,8), (-2,3), (-1,0), (0,-1), (1,0), (2,3), (3,8) on the graph.
*
The Parabola is intersecting the x – axis at the points (-1, 0) and (1,0)
∴ Zeroes of p(x) = -1, 1
(Q3) (i) From the graph,
The parabola is intersecting the x – axis at the points (-3, 0) and (4,0)
∴ Zeroes of p(x) = -3, 4
(ii) From the graph,
The parabola is intersecting the x – axis at the point (3, 0)
∴ Zeroes of p(x) = 3,3
(iii) From the graph,
The parabola is not at all touching or intersecting the x – axis
There are no zeroes
(iv) From the graph,
The parabola is intersecting the x – axis at the points (-4, 0) and (1, 0)
∴ zeroes of p(x) = -4, 1
(Q4) Why are 1/4 and -1 zeroes of the polynomial p(x) = 4x2 + 3x – 1
=> Solution:
Given: P(x) = 4x2 + 3x – 1 ——- (1)
Substituting x = 1/4 in —– (1)
P (1/4) = 4 (1/4)2 +3 (1/4) – 1
P (1/4) = 4 (1/16) + 3/4 – 1
= 1/4 + 3/4 – 1
= 4/4 – 1
= 1 – 1
P (1/4) = 0
∴ 1/4 is a zero of p(x) substituting x = -1 in —— (1)
P (-1) = 4 (-1)2 + 3(-1) -1
= 4 (1) + (-3) – 1
P (-1) = 4 – 3 – 1
P (-1) = 1 – 1
P (-1) = 0
∴ ‘-1’ is zero of p(x)
∴ – 1 and 1/4 are zeroes of p (x)
Here is your solution of Telangana SCERT Class 10 Math Chapter 3 Polynomials Exercise 3.2
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