Mahesh

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

Do this

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

Try This

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

Do This

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.

Try This

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

Do This

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:

Try This

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.

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

Question 1.
Verify that the numbers given along side the cubic polynomials below are their zeroes. Also verify the relationship between the zeros and the coefficients in each case :
i) 2x³ + x² – 5x + 2;(\(\frac{1}{2}\), 1, -2)
ii) x³ + 4x² + 5x – 2; (2, 1, 1)
Solution:
i) Given polynomial is 2x³ + x² – 5x + 2
Comparing the given polynomial with ax³ + bx² + cx + d,
We get a = 2, b = 1, c = -5 and d = 2
p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))³ + (\(\frac{1}{2}\))³ – 5(\(\frac{1}{2}\)) + 2
= \(\frac{1}{4}\) + \(\frac{1}{4}\) – \(\frac{5}{2}\) + \(\frac{2}{1}\)
= \(\frac{1+1-10+8}{4}\) = \(\frac{0}{4}\) = 0
P(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= 2(-8) + 4 + 10 + 2
= -16 + 16 = 0
∴ \(\frac{1}{2}\), 1, and -2 are the zeroes of 2x³ + x² – 5x + 2
So, α = \(\frac{1}{2}\), β = 1 and γ = -2
Therefore,
α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(\frac{1+2-4}{2}\) = \(\frac{-1}{2}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = (\(\frac{1}{2}\))(1) + (1)(-2) + -2(\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1
= \(\frac{1-4-2}{2}\) = \(\frac{-5}{2}\) = \(\frac{c}{a}\)
And αβγ = \(\frac{1}{2}\) × 1 × (-2) = -1 = \(\frac{-2}{2}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

ii) Given polynomial is x³ – 4x² + 5x – 2
Comparing the given polynomial with ax³ + bx² + cx + d, We get a = 1, b = -4, c = 5 and d = -2
p(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
∴ 2, 1 and 1 are the zeroes of x³ – 4x² + 5x – 2.
So, α = 2, β = 1 and γ = 1
Therefore,
α + β + γ = 2 + 1 + 1
= 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = 2(1) + (1)(1) + (1) (2)
= 2 + 1 + 2 = 5
= \(\frac{5}{1}\) = \(\frac{c}{a}\)
and
αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2, -7, -14 respectively.
Answer:
Let the cubic polynomial be
ax³ + bx² + cx + d and its zeroes be α, β and γ.
Then
α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = -7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)
and αβγ = -14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)
∴ a = 1, b = -2, c = -7 and d = 14.
So, one cubic polynomial which satisfies the given conditions will be
x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Answer:
Given polynomial is x³ – 3x² + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b)
⇒ 3a = 3 ⇒ a = 1
= \(\frac{-(-3)}{1}\) = 3
∴ Sum of the products of its zeroes taken two at a time.
= a(a – b) +a(a + b) + (a + b) (a – b)
= \(\frac{1}{1}\) = 1
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
So, 3(1)² – b² = 1
⇒ 3 – b² = 1
⇒ b² = 2 ⇒ b = \(\sqrt{2}\) = ±\(\sqrt{2}\)
Here, a = 1 and b = ±\(\sqrt{2}\)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\) , find other zeroes.
Answer:
Given polynomial is
x⁴ – 6x³ – 26x² + 138x – 35
We have, 2 ± \(\sqrt{3}\) are two zeroes of the polynomial
p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Let x = 2 ± \(\sqrt{3}\)
So, x – 2 = ±\(\sqrt{3}\)
On squaring, we get x² – 4x + 4 = 3, i.e., x² – 4x + 1 = 0
Let us divide p(x) by x² – 4x + 1 to obtain other zeroes

p(x) = x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x² – 7x + 5x – 35)
= (x² – 4x + 1) [x(x – 7) +5(x – 7)]
= (x² – 4x + 1) (x + 5) (x – 7)
So, (x + 5) and (x – 7) are other factors of p(x)
So, -5 and 7 are other zeroes of the given polynomial.

Question 5.
If the polynomial
x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Given polynomial is
x⁴ – 6x³ + 16x² – 25x + 10 and another polynomial is x² – 2x + k
Remainder is x + a
Let us divide
x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k.

∴ Remainder = (2k – 9)x – (8 – k)k + 10
But the remainder is given as x + a.
On comparing their coefficients we have
2k – 9 = 1 ⇒ 2k = 10
⇒ k = 5 and -(8 – k) k + 10 = a
So, a = -(8 – 5) 5 + 10 = -3 × 5 + 10
= -15 + 10 = -5
Hence, k = 5 and a = -5

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.4

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.
i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
Solution:
i) p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
The given polynomials are in standard form.

The degree of x² – 2 is 2.
The degree of 7x – 9 is 1.
∴ We stop here since the degree of (7x – 9) < degree of (x² – 2)
So, the quotient is x – 3 and the remainder is 7x – 9.

ii) p(x) = x⁴ – 3x² + 4x + 5,
g(x) = x² + 1 – x
p(x) = x⁴ – 3x² + 4x + 5, it is in standard form.
g(x) = x² + 1 – x, it is not in standard form. Writing it in standard form, we have x² – x + 1. Now we apply the division algorithm to the given polynomials.

The degree of x² – x + 1 is 2
The degree of 8 is 0.
∴ We stop here since the degree of (8) < degree of (x² – x + 1)
So, the quotient is x² + x – 3 and the remainder is 8.

iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
p(x) = x⁴ – 5x + 6 → it is in standard form
g(x) = 2 – x² → it is not in standard form writing it in standard form, we have -x² + 2.
Now we apply the division algorithm to the given polynomials

The degree of -x² + 2 is 2 and that of (-5x + 10) is 1
The degree of (-x² + 2) > degree of (-5x + 10)
So, we stop here
So, the quotient is -x² – 2 and the remainder is -5x + 1o.

Question 2.
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:
Divide the second polynomial by the first polynomial. If the remainder is zero, then the first polynomial is a factor of the second one. The given polynomials are in standard form.
i)

Since the remainder is ‘0’, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii)

Since the remainder is ‘0’, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii)

Here, the remainder is 2.
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.
x⁵ – 4x³ + x² + 3x + 1

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are
\(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
Solution:
The given polynomial is
3x⁴ + 6x³ – 2x² – 10x – 5.
Two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
∴ [x – \(\sqrt{\frac{5}{3}}\)][x + \(\sqrt{\frac{5}{3}}\)] = [x² – \(\sqrt{\frac{5}{3}}\)] is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial & x² – \(\frac{5}{3}\).

3x⁴ + 6x³ – 2x² – 10x – 5 = (x² – \(\frac{5}{3}\)) (3x² + 6x + 3)
3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)²
So, the zeroes of 3(x + 1)² are -1 and -1.
Therefore, the zeroes of the given polynomial are and \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\), -1 and -1

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
By division algorithm, we have
p(x) = g(x) × q(x) + r(x)
g(x) × q(x) = p(x) – r(x)
g(x) × (x – 2) = x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
∴ g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2)

∴ g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
i) deg p(x) = deg q(x)
ii) deg q(x) = deg r(x)
iii) deg r(x) = 0
Solution:
i) p(x) = 6x² – 12x + 15, g(x) = 3,
q(x) = 2x² – 4x + 5, r(x) = 0
ii) p(x) = x³ + 2x² + x – 6, g(x) = x² + 2
q(x) = x + 2, r(x) = -x – 10
iii) p(x) = x³ + 5x² – 3x – 10, g(x) = x² – 3
q(x) = x + 5, r(x) = 5

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5

Question 1.
Construct ∠ABC = 60° without using protractor.
Answer:

Steps of construction:

1. Draw a line l and choose a point B on it.
2. Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line / at a point say C.
3. Take C as centre and with the same radius (as in step 2) in draw an arc.
4. Now take B as centre and with the same radius (as in step 2), draw another arc cutting the previous arc (drawn in step 2) at A.
5. Join AB, we get ∠ABC whose measure is 60°.

Question 2.
Construct an angle of 120° with using protractor and compasses.
Answer:
With using compasses:

Steps of construction:

1.  Draw any ray OA.
2. Place the pointer of the compasses at O with O as centre and any convenient radius draw an arc cutting OA at M.
3. With M as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at P.
4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
5. Join OQ. Then ∠AOQ is the required angle.

With Using protractor:

Steps of construction :

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at
   Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark at a point P at 120°.
4. Join QP. ∠PQR is the required angle.

Question 3.
Construct the following angles using ruler and compasses. Write the steps of construction in each case.
(i) 75°
Answer:

Steps of construction:

1. Draw any ray OA.
2. Place the pointer of the compasses at O. With ‘O’ as centre and any convenient radius draw an arc cutting OA at M.
3. With M as centre and without altering radius (as in step 2) draw an arc which, cuts the first arc at P.
4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
5. Join OQ. ∠AOQ = 120° (∵∠AOP + ∠POQ = 60°)
   ∠AOQ = ∠AOP + ∠POQ = 60° + 60° = 120°
6. Draw OB the perpendicular bisector of ∠POQ.
   Now ∠POB = ∠QOB = 30°
   ∴∠AOB = 90°
7. Draw OC the perpendicular bisector of ∠POB.
   Now ∠POC = ∠BOC = 15°
8. ∠AOC = ∠AOP + ∠OC
   = 60° + 15°
   = 75°
   (or)
   ∠AOC = ∠AOB – ∠BOC
   = 90° – 15°
   = 75°
   ∠AOC is the required angle whose measure is 75°.

(ii) 15°
Answer:

Steps of construction:

1. Draw a line l and mark a point O on it.
2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line / at a point say A.
3. With the pointer at A (as centre) and the same radius as in the step – 2, now draw an arc that passes through O.
4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
5. Draw the bisector OC of ∠AOB. Now ∠AOC = 30°.
   [∠AOC = ∠BOC = 30° ]
6. Draw the bisector OD of ∠AOC. Now ∠AOD = 15°.
   [∠AOD = ∠COD = 15°]
7. ∠AOD = 15° is the required angle.

(iii) 105°
Answer:

Steps of construction:

1. Draw any ray OM.
2. Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OM at A.
3. With A as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
4. With B as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
5. Join QC. ∠AOC = 120°.
6. We know that ∠AOB = ∠BOC = \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\) × 120° = 60°
7. Draw OD the bisector of ∠BOC.
   ∴ ∠BOD = ∠DOC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) × 60° = 30°
8. Draw OE the bisector of ∠DOC.
   ∴ ∠DOE = ∠EOC = \(\frac{1}{2}\) ∠DOC = \(\frac{1}{2}\) × 30° = 15°
9. Now ∠AOE = ∠AOD + ∠DOE = 90° + 15° = 105°
10. ∠AOE = 105° is the required angle.

Question 4.
Draw the angles given in Q. 3 using a protractor.
Answer:
The angles given in Q. 3 are 75°, 105° and 15°.

(i) 75°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 75°.
4. Join QP. ∠RQP is the required angle.

(ii) 105°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 105°
4. Join QP. ∠RQP is the required angle.

(iii) 15°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 15°
4. Join QP. ∠RQP is the required angle.

Question 5.
Construct ∠ABC = 50° and then draw another angle ∠XYZ equal to ∠ABC without using a protactor.
Answer:

Steps of construction:

1. Make an angle ∠ABC = 50° v
2. Draw a line l and choose a point Y on it.
3. Use the same compasses setting to draw an arc with Y as centre, cutting l in Z.
4. Set your compasses to the length MN.
5. With the same radius place the compasses pointer at Z and draw an arc to cut the arc drawn earlier at X.
6. Join XY. This gives us ∠XYZ. It has the same measure as ∠ABC i.e,, 50°.

Question 6.
Construct ∠DEF = 60°. Bisect it, measure each half by using a protractor.
Answer:

Steps of construction :

1. Draw a line P and mark a point ‘E’ on it.
2. Place the pointer of the compasses at E and draw an arc of convenient radius which cuts the line \(\overleftrightarrow{\mathrm{PQ}}\) at a point say F.
3. With the pointer at F (as centre), now draw an arc that passes through E.
4. Let the two arcs intersect at D. Join ED. We get ∠DEF whose measure is 60°.
5. Draw OC the bisector of ∠DEF.
6. Now ∠FEC = ∠CED = 30°.

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.4

Question 1.
Draw the following angles with the help of a protractor.
(i) ∠ABC = 65°
Answer:
∠ABC = 65°

Steps of construction:

1. Draw a ray BC of any length.
2. Place the centre point of the protractor at B and the line aligned with the \(\overrightarrow{\mathrm{BC}}\).
3. Mark a point A at 65°.
4. Join BA. ∠ABC is the required angle.

(ii) ∠PQR = 136°
Answer:
∠ PQR = 136°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overrightarrow{\mathrm{QR}}\).
3. Mark a point P at 136°.
4. Join QP. ∠PQR is the required angle.

(iii) ∠Y = 45°
Answer:
∠Y = 45°

Steps of construction:

1. Draw a ray YZ of any length.
2. Place the centre point of the protractor
   at Y and the line aligned with the \(\overrightarrow{\mathrm{YZ}}\).
3. Mark a point X at 45°.
4. Join YX. ∠XYZ is the required angle.

(iv) ∠O =172°
Answer:
∠O = 172°

Steps of construction:

1. Draw a ray OP of any length.
2. Place the centre point of the protractor at 0 and the line aligned with the OP
3. Mark a point N at 172°.
4. Join ON. ∠NOP is the required angle.

Question 2.
Copy the following angles in your notebook and find their bisector.

Answer:

Steps of construction :

1. Draw a line l and choose a point P on it.
2. Now place the compasses at A and draw an arc to cut the rays AC and AB.
3. Use the same compasses setting to draw an arc with P as centre, cutting / at Q.
4. Set your compasses with \(\overline{\mathrm{BC}}\) as the radius.
5. Place the compasses pointer at Q and draw an arc to cut the existing arc at R.
6. Join PR. This gives us ∠RPQ. It has the same measure as ∠CAB.
   This means ∠QPR has same measure as ∠BAC.
7. Take Q as centre and with radius more than half of the length PQ, draw an arc. Now take R as centre and with the same radius, draw another arc intersecting the previous arc at S. Join PS. PS is the bisector of the given angle.

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.3

Question 1.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.

i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Solution:
i) Let p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The zeroes of p(x) are given by P(x) = 0
⇒ (x + 2) (x – 4) = 0
⇒ x + 2 = 0 (or) x – 4 = 0
⇒ x = -2 (or) x = 4
Hence the zeroes of x² – 2x – 8 are -2 and 4.
Sum of the zeroes = -2 + 4 = 2

Product of the zeroes = -2 × 4 = -8

ii) Let p(s) = 4s² – 4s + 1
= (2s)² – 2(2s) (1) + (1)²
= (2s – 1)²
= (2s – 1) (2s – 1)
The zeroes of p(s) are given by p(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 =0 (or) 2s – 1 = 0
⇒ s’ = \(\frac{1}{2}\) (or) s = \(\frac{1}{2}\)
Hence the zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\), \(\frac{1}{2}\)
Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1;

(iii) Let p(x) = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
The zeroes of p(x) are given by p(x) = 0
⇒ (3x + 1) (2x – 3) = 0
⇒ 3x + 1 = 0 (or) 2x – 3 = 0
⇒ 3x = -1 (or) 2x = 3
⇒ x = \(\frac{-1}{3}\) (or) x = \(\frac{3}{2}\)
Hence the zeroes of 6x – 7x – 3 are \(\frac{-1}{3}\) and \(\frac{3}{2}\)
Sum of the zeroes = \(\frac{-1}{3}\) + \(\frac{3}{2}\)

iv) Let p(u) = 4u² + 8u
= 4u(u + 2)
The zeroes of p(u) are given by
⇒ p(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u = 0 (or) u + 2 = 0
⇒ u = 0 (or) u = -2
Hence the zeroes of 4u + 8u are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2

v) Let p(t) = t² – 15
The zero of p(t) is given by p(t) = 0
⇒ t² – 15 = 0 ⇒ t² = 15

vi) Let P(x) = 3x²
The zero of p(x) is given by P(x) = 0
⇒ 3x² – x – 4 = 0
⇒ 3x² + 3x – 4x – 4 = 0
⇒ 3x(x + 1) – 4(x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0 (or) 3x – 4 = 0
⇒ x = -1 (or) x = 4/3
Hence the zeroes of 3x² – x – 4 are -1 (or) 4/3
Sum of the zeroes of 3x² – x – 4 are -1 (or) 4/3.
Sum of the zeroes =

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

i) \(\frac{1}{4}\), -1
ii) \(\sqrt{2}\), \(\frac{1}{3}\)
iii) 0, \(\sqrt{5}\)
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1 (A.P. Mar. ’15)
Solution:
i) \(\frac{1}{4}\), -1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\frac{1}{4}\)
Product of the zeroes = αβ = -1
The required quadratic polynomial will be
k[x² – x (α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\frac{1}{4}\)) + (-1)]
⇒ k(x² – \(\frac{x}{4}\) – 1)
If k = 4, then the polynomial will be 4(x² – \(\frac{x}{4}\) – 1) = 4x² – x – 4

ii) \(\sqrt{2}\), \(\frac{1}{3}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\sqrt{2}\)
Product of the zeroes = αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\sqrt{2}\)) + \(\frac{1}{3}\)]
⇒ k[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)]
when k = 3, then the polynomial will be
3[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)] (or) 3x² – 3\(\sqrt{2}\)x + 1

iii) 0, \(\sqrt{5}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = \(\sqrt{5}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ where k is a constant
⇒ k[x² -x(0) + \(\sqrt{5}\)]
⇒ k[x² + \(\sqrt{5}\)].
Where k = 1, the polynomial will be x² + \(\sqrt{5}\).

iv) 1, 1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 1
Product of the zeroes = αβ = 1
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(α – β) + 1]
⇒ k[x² – x + 1]
Where k = 1, the polynomial will be x² – x + 1.

v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = –\(\frac{1}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(-\(\frac{1}{4}\)) + \(\frac{1}{4}\)]
⇒ k[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)]
Where k = 4, the polynomial will be
4[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)] = 4x² + x + 1

Question 3.
Find the quadratic polynomial for the zeroes α, β given in each case. (A.P.Mar.’16)

i) 2, -1
ii) \(\sqrt{3}\), –\(\sqrt{3}\)
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Solution:
i) 2,-1
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = 2, β = -1
Sum of the zeroes = α + β = 2 + (-1) = 1
Product of the zeros = αβ = 2 × (-1)
= -2
∴ The required quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(1) + (-2)]
⇒ k[x² – x – 2]
Where k = 1, the quadratic polynomial will be x² – x – 2.

ii) \(\sqrt{3}\), –\(\sqrt{3}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\sqrt{3}\) , β = –\(\sqrt{3}\)
Sum of the zeroes = α + β
= \(\sqrt{3}\) + (-\(\sqrt{3}\)) = o
Product of the zeros
= αβ = \(\sqrt{3}\) × –\(\sqrt{3}\) = -3
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(0) + (-3)]
⇒ k[x² – 3]
Where k = 1, the quadratic polynomial will be [x² – 3].

iii) \(\frac{1}{4}\), -1
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{4}\), β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{-3}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\) × (-1)
= –\(\frac{1}{4}\)
Therefore, the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
Where k = 4, the quadratic polynomial will be [4x² + 3x – 1]

iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{2}\), β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant

Where k = 4, the quadratic polynomial will be [4x² – 8x + 3].

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients. (A.P. Mar. ’15)
Solution:
The given polynomial is x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d
We get a = 1, b = 3, c = -1, d = -3
Let p(x) = x³ + 3x² – x – 3
Then p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3
= 4 – 4 = 0
p(-1) = (-1)³ + 3(-1)² – (-1) – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore 1, -1 and -3 are the zeroes of
x³ + 3x² – x – 3.
So α = 1, β = -1, γ = -3
α + β + γ = 1 – 1 – 3 = -3 = -3/1 = \(\frac{-b}{a}\)
αβ + βγ + αγ = 1(-1) + (-1)(-3) + (-3)(1)
= -1 + 3 – 3 = -1 = -1/1 = c/a
αβγ = (1)(-1)(-3) = – \(\left(\frac{-3}{1}\right)\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Try These

Question 1.
Can you now write the rule to form the following pattern with match-sticks ?

Answer:
The rule for the above figures is 3n.
[∵ 3 × 1, 3 × 2, 3 × 3, ………………]

Question 2.
Find the rule for required number of matchsticks to form a pattern repeating ‘H’. How would the rule be for repeating the shape ‘L’ ?
Answer:
The required pattern of repeating ‘H’ is
‘3n’. (3 × 1, 3 × 2, ………………..)
The rule for ‘L’ is ‘2n’.

Question 3.
A line of shapes is constructed using matchsticks.

(i) Find the rule that shows how many sticks are needed to make a group of such shapes ?
(ii) How many matchsticks are needed to form a group of 12 shapes ?
Answer:
(i) Number of matchsticks that are used in each figure are
3 ; 5 ; 7 ; 9 ………………
Shape-1; Shape-2; Shape-3; Shape-4
2 × 1 + 1; 2 × 2 + 1 ; 2 × 3 + 1 ; 2 × 4 + 1 ………
In general (2n + 1) is the rule for the above shapes.

(ii) 2n + 1 = 2 × 12 + 1 = 25 [ ∵ n = 12]

(Try These)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables ‘l’ and ‘b’ for length and breadth of the rectangle respectively.
Answer:

Perimeter of a rectangle
= l + b + l + b
= 2l + 2b = 2 (l + b)

Question 2.
Find the general rule for the area of a square by using the variable ’s’ for the side of a square ?
Answer:

The rule for the area of a square is A = side × side
= s × s
A = s²

Question 3.
What would be the rule for perimeter of an Isosceles triangle ?
Answer:

The rule for the perimeter of an Isosceles triangle
= 3 × side
= 3 × a
P = 3a

Do This

Question 1.
Find the n^th term in the following sequences.
(i) 3, 6, 9, 12, …………………
(ii) 2, 5, 8, 11, …………………
(iii) 1, 8, 27, 64, 125, …………………
Answer:
(i) The n^th> term of the above series is
3 × 1, 3 × 2, 3 × 3, ……………….. = 3n.

(ii) The n^th term is 3 × 1 – 1, 3 × 2 – 1, 3 × 3 – 1, 3 × 4 – 1,
i.e., 3n – 1

(iii) The nth term is 1 × 1 × 1, 2 × 2 × 2, 3 × 3 × 3, ……. is
1³, 2³, 3³ ………………….
i.e., n³

Question 2.
Write LHS and RHS of the following simple equations:
(i) 2x + 1 = 10
Answer:
2x + 1 = 10
LHS = 2x + 1, RHS = 10

(ii) 9 = y – 2
Answer:
9 = y – 2 ;
LHS = 9, RHS = y – 2

(iii) 3p + 5 = 2p + 10
Answer:
3p + 5 = 2p + 10
LHS = 3p + 5, RHS = 2p + 10

Question 3.
Write any two simple equations and give their LHS and RHS.
Answer:

Question 4.
Find the solution of the equation ‘x – 4 = 2’ by Trail and Error method.
Answer:
Given equation is x – 4 = 2
If x = 1 ⇒ x – 4 = 2
⇒ 1 – 4 = 2
– 3 ≠ 2
If x = 3 ⇒ 3 – 4 = 2
– 1 ≠ 2
If x = 5 ⇒ 5 – 4 = 2 .
1 ≠ 2
If x = 6 ⇒ 6 – 4 = 2
2 = 2
x = 6 satisfies the given equation.
∴ x = 6 is the solution of the equation.

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.2

Question 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 zeroes
of p(x).
Solution:

Solution:
Six figures are given and they are numbered.
In the first figure, there are no zeroes.
In the second figure, there is one zero.
In the third figure, there are three zeroes.
In the fourth figure, there are two zeroes.
In the fifth figure, there are four zeroes.
In the sixth figure, there are three zeroes.

Question 2.
Find the zeroes of the given polynomials.
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Solution:
i) p(x) = 3x
p(0) = 3 × 0 = 0
∴ The zero of p(x) = 3x is 0

ii) p(x) = x² + 5x + 6
= x² + 3x + 2x + 6
= x(x + 3) + 2(x + 3) = (x + 2) (x + 3)
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
Therefore, the zeroes of x + 5x + 6 are – 2 and -3.

iii) p(x) = (x + 2) (x + 3)
⇒ x² + 2x + 3x + 6
⇒ x² + 5x + 6
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
∴ The zeroes of (x + 2) (x + 3) are -2 and -3.

iv) p(x) = x⁴ – 16
= (x²)² – (4)² = (x² + 4) (x² – 4)
= (x² + 4)(x + 2)(x – 2)
To find zeroes, let p(x) = O
⇒ (x² + 4) (x + 2)(x – 2) = 0
(i.e.,) x² + 4 = 0 (or) x + 2 = 0 (or) x – 2 = 0
If x² + 4 = 0, then x² = -4 ⇒
x = ± \(\sqrt{-4}\)
If x + 2 = 0, then x = -2
If x – 2 = 0, then x = 2
∴ The zeroes of the polynomial x⁴ – 16 are -2, 2 and ± \(\sqrt{-4}\).

Question 3.
Draw the graphs of the given polynomials and find the zeroes. Justify the answers. (A.P.June’ 15)
(i) p(x) = x² – x – 12
(ii) p(x) = x² – 6x + 9
(iii) p(x) = x² – 4x + 5
(iv) p(x) = x² + 3x – 4
(v) p(x) = x² – 1
Solution:
i) p(x) = x² – x – 12

∴ The zeros of x² – x – 12 are 4 and -3.

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph intersects the X-axis at (4, 0) and (-3, 0)

ii) p(x) = x² – 6x + 9


The graph intersects one the X-axis at only point (3, 0)

iii) p(x) = x² – 4x + 5

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph does not intersect at the X-axis.
There are no zeroes of the polynomial x² – 4x + 5

iv) p(x) = x² + 3x – 4


∴ The graph intersects the X-axis at (1, 0) and (-4, 0)
∴ The zeroes of the polynomial x² + 3x – 4 are 1 and -4.

v) p(x) = x² – 1


The graph intersects the X-axis at (-1, 0) and (1, 0)
∴ The zeroes of the polynomial x² – 1 are -1 and 1.

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1?
Solution:
p(x) = 4x² + 3x – 1
∴ p(\(\frac{1}{4}\)) = 4(\(\frac{1}{4}\))² + 3(\(\frac{1}{4}\)) – 1
= \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{1}{1}\)
= \(\frac{1+3-4}{4}\) = \(\frac{0}{4}\) = 0
∴ p(-1) = 4(-1)² + 3(-1) – 1
= 4(1) – 3 – 1
= 4 – 3 – 1
= 4 – 4 = 0
Since p(\(\frac{1}{4}\)) and p(-1) are equal to zero.
\(\frac{1}{4}\) and -1 are the zeroes of the polynomial.