Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

Solution:

9th Maths Exercise 3.1 Solutions Question 3.
Find the degree of the following polynomials.

Solution:

Samacheer Kalvi Guru 9th Maths Question 4.
Rewrite the following polynomial in standard form.

Solution:

9th Standard Maths Algebra Question 5.
Add the following polynomials and find the degree of the resultant polynomial.
(i) p(x) = 6x² – 7x+ 2 q(x) = 6x³ – 7x + 15
(ii) h(x) = 7x³ – 6x + 1 f(x) = 7x² + 17x – 9
(iii) f(x) = 16x⁴ – 5x² + 9 g(x) = -6x³ + 7x – 15
Solution:

9th Standard Maths Exercise 3.1 Question 6.
Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x² + 6x – 1 q(x) = 6x – 9
(ii) f(y) = 6y² – 7y + 2 g(y) = 7y + y³
(iii) h(z) = z⁵ – 6z⁴ + z f(z) = 6z² + 10z – 7
Solution:

Samacheer Kalvi Guru 9 Maths Question 7.
What should be added to 2x³ + 6x² – 5x + 8 to get 3x³ – 2x² + 6x + 15?
Solution:
(2x³ + 6x² – 5x + 8) + Q(x) = 3x³ – 2x² + 6x + 15
∴ Q(x) = (3x³ – 2x² + 6x + 15) – (2x³ + 6x² – 5x + 8)

The required polynomial is x³ – 8x² + 11x + 7

Algebra 9th Standard Question 8.
What must be subtracted from 2x²⁴ + 4x² – 3x + 7 to get 3x³ – x² + 2x + 1 ?
Solution:
(2x⁴ + 4x² – 3x + 7) – Q(x) = 3x³ – x² + 2x + 1
Q(x) = (2x⁴ + 4x² – 3x + 7) – 3x³ – x² + 2x + 1
The required polynomial = 2x⁴ + 4x² – 3x + 7 – 3x³ + x² – 2x – 1
= 2x⁴ – 3x³ + 5x² – 5x + 6

Ex 3.1 Class 9th Maths Solution Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x² – 9 q(x) = 6x² + 7x – 2
(ii) f(x) = 7x + 2 g(x) = 15x – 9
(iii) h(x) = 6x² – 7x + 1 f(x) = 5x – 7
Solution:
(i) p(x) = x² – 9 q(x) = 6x² + 7x – 2
p(x) × q(x) = (x² – 9) (6x² + 7x – 2)

The required polynomial is 6x⁴ + 7x³ – 56x² – 63x + 18, degree 4.

(ii) f(x) = 7x + 2 g(x) = 15x – 9
f(x) × g(x) = (7x + 2) (15x – 9)

The required polynomial is 105x² – 33x – 18, degree 2.

(iii) h(x) = 6x² – 7x + 1 f(x) = 5x – 7
h(x) × f(x) = (6x² – 7x + 1) (5x – 7)

The required polynomial is 30x³ – 77x² + 54x – 7, degree 3.

9th Class Math Exercise 3.1 Solution Question 10.
The cost of chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x andy. If x = 10, y = 5 find the amount paid by him.
Solution:
Amount paid = Number of chocolates × Cost of a chocolate
= (x + y) (x + y) = (x + y)² = x² + 2xy + y²
If x = 10, y = 5
The total amount paid by him
= 10² + 2 × 10 × 5 + 5² = 100 + 100 + 25 = Rs. 225

Samacheer Kalvi Guru 9th Maths Guide Question 11.
The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution:
Area of a rectangle = length × breadth
= (3x + 2) × (3x – 2) = (3x)² – 2² = [9x² – 4] Sq. units
If x = 20, Area = 9 × 20² – 4 = 9 × 400 – 4
= 3600 – 4 = 3596 Sq. units

Samacheer Guru 9th Maths Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial p(x) × q(x) is ?
Solution:
p(x) is a polynomial of degree 1. q(x) is a polynomial of degree 2.
Then the p(x) × q(x) will be the polynomial of degree (1 + 2) = 3 (or)
Cubic polynomial