TS 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

Students can practice 10th Class Maths Solutions Telangana Chapter 3 Polynomials InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials InText Questions

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Question 1.
State which of the following are polynomials and which are not ? Give reasons. (Page No. 48)
(i) 2x³
(ii) \(\frac{1}{x-1}\)
(iii) 4z² + \(\frac{1}{7}\)
(iv) m² – \(\sqrt{2}\) m + 2
(v) p^-2 + 1
Solution:
i) 2x³ is a polynomial
ii) \(\frac{1}{x-1}\) is not a polynomial.
iii) 4z² + \(\frac{1}{7}\) is a polynomial.
iv) m² – \(\sqrt{2}\)m + 2 is a polynomial.
v) p^-2 + 1 is not a polynomial.

Question 2.
p(x) = x² – 5x – 6, find the values of p(1), p(2), p(3), p(0), p(-1), p(-2), p(-3). (Page No. 49)
Solution:
p(x) = x² – 5x – 6
p(1) = (1)² – 5(1) – 6 = 1 – 5 – 6 = 1 – 11 = -10
p(2) = (2)² – 5(2) – 6 = 4 – 10 – 6
= 4 – 16 = – 12
p(3) = (3)² – 5(3) – 6 = 9 – 15 – 6
= 9 – 21 = -12
p(0) = (0)² – 5(0) – 6 = 0 – 0 – 6 = 0 – 6 = -6
p(-1) = (-1)² – 5(-1) – 6 = 1 + 5 – 6 = 6 – 6 = 0
p(-2) = (-2)² – 5(-2) – 6 = 4 + 10 – 6 = 8
p(-3) = (-3)² – 5(-3) – 6 = 9 + 15 – 6
= 24 – 6 = 18

Question 3.
p(m) = m² – 3m + 1, find the value of p(1) and p(-1). (Page No. 49)
Solution:
p(m) = m² – 3m + 1
p(1) = (1)² – 3(1) + 1 = 1 – 3 + 1 = 2 – 3 = -1
p(-1) = (-1)² – 3(-1) + 1 = 1 + 3 + 1 = 5

Question 4.
Let p(x) = x² – 4x + 3. Find the value of p(0), p(1), p(2), p(3) and obtain zeroes of the polynomial p(x).
(Page No. 50)
Solution:
p(x) = x² – 4x + 3
p(0) = (0)² – 4(0) + 3 = 0 – 0 + 3 = 3
p(1) = (1)² – 4(1) + 3 = 1 – 4 + 3 = 4 – 4 = 0
p(2) = (2)² – 4(2) + 3 = 4 – 8 + 3 = 7 – 8 = -1
p(3) = (3)² – 4(3) + 3 = 9 – 12 + 3 = 12 – 12 = 0
We see that p(1) = 0 & p(3) = 0
These points x = 1 & x = 3 are called zeroes of the polynomial p(x) = x² – 4x + 3

Question 5.
Check whether -3 and 3 are the zeroes of the polynomial x² – 9. (Page No. 50)
Solution:
p(x) = x² – 9 ⇒ p(-3) = (-3)² – 9 = 9 – 9 = 0
p(3) = (3)² – 9 = 9 – 9 = 0
p(-3) = 0 & p(3) = 0
∴ -3 and 3 are the zeroes of the polynomial p(x) = x² – 9

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Question 1.
Write 3 different quadratic, cubic and 2 linear polynomials with different number of terms. (Page No. 48)
Solution:
3 different quadratic polynomials :

1. 3x² + 2x + 4
2. 5x² + x + 1
3. x² – 3x + 2

3 different cubic polynomials :

1. x³ + 2x² + x + 1
2. 3x³ + 2x + 1
3. x³ + 1

2 linear polynomials :

1. 4x + 3
2. 3x + 2

Question 2.
Write a quadratic polynomial and a cubic polynomial in variable x in the general form. (Page No. 49)
Solution:
General form of a (in variable x)

1. Quadratic polynomial → ax² + bx +c.
2. Cubic polynomial → ax³ + bx² + cx + d

Question 3.
Write a general polynomial q(z) of degree n with coefficients that are b₀,…,b_n. What are the conditions on b₀, ..,b_n ? (Page No. 49)
Solution:
General polynomial q(z) of degree ‘n’ is q(z) = b₀xⁿ + b₁x^n-1 + b₂x^n-2 + b₃x^n-3 +…. +b_n-1x + b_n

Conditions :

1. b₀, b₁, b₂, …….,b_n-1, b_n are real coefficients.
2. bg ≠ 0

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Question 1.
Draw the graph of
(i) y = 2x + 5
(ii) y = 2x – 5
(iii) y = 2x and find the point of intersection on X-axis. Is the x- coordinate of these points also the zero of the polynomial ? (Page No. 52)

i) y = 2x +


The zero of the polynomial 2x + 5 is the x-coordinate of the point where the graph of y = 2x + 5 intersects the X-axis.
∴ Zero of the polynomial 2x + 5 is \(\frac{-5}{2}\).

ii) y = 2x – 5

The zero of the polynomial 2x – 5 is the x-coordinate of the point where the graph of y = 2x – 5 intersects the X-axis.
∴ Zero of the polynomial 2x – 5 is 5/2.

iii) y = 2x

The zero of the polynomial 2x is the x-coordinate of the point where the graph of y = 2x intersects the X-axis.
The graph y = 2x does not intersect X-axis.
∴ There is no zero of the polynomial for 2x.

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Question 1.
Draw the graphs of
(i) y = x² – x – 6
(ii) y = 6 – x – x² and find zeroes in each case. What do you notice ? (A.P. Mar. 15) (Page No. 53)
Solution:
i) y = x² – x – 6

Zeroes of the quadratic polynomial x² – x – 6 are the x-co-ordinates of the points of X-axis.

-2 & 3 are zeroes of the quadratic polynomial and -2 & 3 are intersections points of X-axis.

ii) y = 6 – x – x²

Zero of the quadratic polynomial 6 – x – x² are the x-coordinates of the points of X-axis.
-3 & 2 are zeroes of the quadratic polynomial and -3 & 2 are intersection points of X-axis.

Observations : The graph cuts X-axis at two distinct points.
The parabola can open either upward or down-ward.

Question 2.
Write three quadratic polynomials that have 2 zeroes each. (Page No. 55)
Answer:

1. y = x² + 5x + 6
2. y = x² – 5x + 6
3. y = 2x² + x – 1

The above three polynomials have 2 zeroes each.

Question 3.
Write one quadratic polynomial that has one zero. (Page No. 55)
Answer:
y = x + 2 this polynomial has one zero.

Question 4.
How will you verify if a quadratic polynomial it has only one zero ?(Page No. 55)
Answer:
The linear polynomial ax + b, a ≠ 0 has exactly one zero namely the x co-ordinate of the point where the graph of y = ax + b intersects X-axis. Every linear polynomial in ax + b, a ≠ 0 has exactly one zero.

Question 5.
Write three quadratic polynomials that have no zeroes for x that are real numbers. (Page No. 55)
Answer:

1. y = 8x² – 6x + 1
2. y = x² – 8x + 17
3. y = x² – 4x + 5

These polynomials have no zeroes for x.

Question 6.
Find the zeroes of cubic polynomials
(i) -x³
(ii) x² – x³
(iii) x³ – 5x² + 6x without drawing the graph of the polynomial. (Page No. 57)
Solution:
i) p(x) = -x³.
p(x) = 0
-x³ = 0
x³ = 0
x = 0. So ‘0’ is the zero of the polynomial -x³.

ii) p(x) = x² – x³
= x²(1 – x)
p(x) = 0
x²(1 – x) = 0 ⇒ x² = 0 or 1 – x = 0
x = 0 (or) x = 1
0 & 1 are the zeroes of the polynomial x² – x³

iii) p(x) = x³ – 5x² + 6x
= x(x² – 5x + 6)
= x(x² – 3x – 2x + 6)
= x[x(x-3) – 2(x – 3)]
= x[(x – 3) (x – 2)]
p(x) =0
x[(x – 3) (x – 2)] = 0 ⇒ x = 0, x – 3 = 0
⇒ x = 3, x – 2 = 0 ⇒ x = 2.
∴ 0, 2 and 3 are the zeroes of the polynomial x³ – 5x² + 6x

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Question 1.
Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial .
(i) p(x) = x² – x – 6
(ii) p(x) = x² – 4x + 3
(iii) p(x) = x² – 4
(iv) p(x) = x² + 2x + 1. (Page No. 62)
Solution:
i) The given polynomial is x² – x – 6.
x² – x – 6 = x² – 3x + 2x – 6
= x(x – 3) +2(x – 3)
= (x + 2) (x – 3)
So, the value of x² – x – 6 is zero when x + 2 = 0 or x – 3 = 0 (i.e.,) when x = -2 or x = 3
Therefore, the zeroes of x² – x – 6 are -2 & 3
Now, sum of the zeroes = -2 + 3 = 1

Product of the zeroes = -2 × (3) = -6

ii) The given polynomial is x² – 4x + 3. (A.P. Mar. June 15)
x² – 4x + 3 = x² – 3x – x + 3
= x(x – 3) -1(x – 3)
= (x – 1) (x – 3)
So, the value of x² – 4x + 3 is zero when (x – 1) = 0 or (x – 3) = 0
(i.e.,) when x = 1 or x = 3
∴ The zeroes of x – 4x + 3 are 1 & 3.
Now sum of zeroes = 1 + 3 = 4

Product of the zeroes = 1 × 3 = 3

iii) The given polynomial is x² – 4
x² – 4 = (x)² – (2)² = (x + 2) (x – 2)
So, the value of x² – 4 is zero when (x + 2) = 0 or (x – 2) = 0
(i.e.,) when x = -2 or x = 2
∴ The zeroes of x² – 4 are -2 & 2.
Now, sum of the zeroes = -2 + 2 = 0

Product of the zeroes = (-2) × (2) = – 4

iv) The given polynomial is x² + 2x + 1.
x² + 2x + 1 = x² + x + x + 1
= x(x +1) +1 (x + 1)
= (x + 1) (x + 1)
= (x + 1)²
So, the value of x² + 2x + 1 is zero when (x + 1) = 0 = x + 1 = 0 (i.e.,) when x = -1 & x = -1.
∴ The zeroes of the polynomial are -1 & -1. Now, sum of the zeroes = -1 + (-1)
= -1 -1 = – 2

Question 2.
If α, β, γ are the zeroes of the given cubic polynomials, find the values as given in the table. (Page No. 66)

Solution:

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Question 1.
i) Find a quadratic polynomial with zeroes -2 and 1/3. (Page No. 64)
Solution:
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 & its zeroes be α & β.
Here α = -2; β = 1/3
sum of the zeroes = (α + β)
sum of the zeros = (α + β)
= (-2) + \(\frac{1}{3}\) = \(\frac{-6+1}{3}\) = –\(\frac{5}{3}\)
Product of the zeroes = αβ = (-2)(\(\frac{1}{3}\))
= –\(\frac{2}{3}\)
∴ The quadratic polynomial
ax² + bx + c is
k[x² – (α + β)x + αβ] where k is a constant.
= k[x² – \(\left(\frac{-5}{3}\right)\)x + \(\left(\frac{-2}{3}\right)\)]
We can put different values of k.
When k = 3 the quadratic polynomial will be 3x² + 5x – 2.

ii) What is the quadratic polynomial whose sum of zeroes is \(\frac{-3}{2}\) and the product of zeroes is -1 ? (Page No. 64) (A.P. June’15)
Solution:
Sum of zeroes = (α + β) = \(\frac{-3}{2}\)
Product of zeroes = αβ = -1
∴ The quadratic polynomial
ax² + bx + c is k[x² – (α + β)x + αβ]
We can put different values of k.
When k = 2 the quadratic polynomial will be 2x² + 3x – 2.