Samacheer Kalvi 9th Maths Book Solutions Chapter 3 – Algebra is available here. The Samacheer Kalvi 9th Maths book answers of Chapter 3, available at BYJU’S, contain step by step explanations designed by our mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 9th Maths solutions. Students can easily refer to these Samacheer Kalvi Class 9 Maths Chapter 3 Questions and Solutions for practise. Doing so will help them score well in exams.
Samacheer Kalvi Class 9 Maths Chapter 3 Questions and Solutions
Chapter 3 of the Samacheer Kalvi 9th Maths guide will help the students to solve problems related to the constants, variables, algebraic expressions, coefficients, polynomials, types of polynomials, the arithmetic of polynomials, value and zeros of a polynomial, roots of a polynomial equation, remainder theorem, factor theorem, algebraic identities, identities involving the product of three binomials, factorisation and its types, division of polynomials, synthetic division, greatest common divisor, to find GCD by factorisation, linear equation in two variables, simultaneous linear equations, methods to solve simultaneous linear equations.
Samacheer Kalvi 9th Maths Chapter 3: Algebra Book Exercise 3.1 Questions and Solutions
Question 1: Which of the following expressions are polynomials? If not give reason:
(i) (1 / x2) + 3x – 4
(ii) x2 (x – 1)
(iii) (1 / x) (x + 5)
(iv) (1 / x-2) + ( 1 / x-1) + 7
(v) √5 x2 + √3 x + √2
(vi) m2 – ∛m + 7m – 10
Solution:
| Given Expression | Is it a polynomial? | Reason |
| (i) (1 / x2) + 3x – 4 = x-2 + 3x – 4 | No | Negative integral power |
| (ii) x2 (x – 1) | Yes | Positive integral power |
| (iii) (1 / x) (x + 5) = x-1 * (x + 5)
= x-1+1 * 5x-1 = x0 + 5x-1 |
No | One of the integral powers is negative. |
| (iv) (1 / x-2) + ( 1 / x-1) + 7 = x2 + x + 7 | Yes | Positive integral power |
| (v) √5 x2 + √3 x + √2 | Yes | Positive integral power |
| (vi) m2 – ∛m + 7m – 10
= m2 – m1/3 + 7m – 10 |
No | One of the powers is fractional. |
Question 2: Write the coefficient of x2 and x in each of the following polynomials.
(i) 4 + (2 / 5)x2 – 3x
(ii) 6 – 2x2 + 3x3 – √7 x
(iii) πx2 – x + 2
(iv) √3 x2 + √2x + 0.5
(v) x2 – (7 / 2)x + 8
Solution:
| Given Expression | Coefficient of x2 | Coefficient of x |
| (i) 4 + (2 / 5)x2 – 3x | 2 / 5 | -3 |
| (ii) 6 – 2x2 + 3x3 – √7 x | -2 | – √7 |
| (iii) πx2 – x + 2 | π | -1 |
| (iv) √3 x2 + √2x + 0.5 | √3 | √2 |
| (v) x2 – (7 / 2)x + 8 | 1 | – (7 / 2) |
Question 3: Find the degree of the following polynomials.
(i) 1 – √2y2 + y7
(ii) (x3 – x4 + 6x6) / x2
(iii) x3 (x2 + x)
(iv) 3x4 + 9x2 + 27x6
(v) 2√5p4 – (8p3 / √3) + (2p2 / 7)
Solution:
| Given Expression | Degree of the polynomial |
| (i) 1 – √2y2 + y7 | 7 |
| (ii) (x3 – x4 + 6x6) / x2 | 4 |
| (iii) x3 (x2 + x) | 5 |
| (iv) 3x4 + 9x2 + 27x6 | 6 |
| (v) 2√5p4 – (8p3 / √3) + (2p2 / 7) | 4 |
Question 4: Rewrite the following polynomial in standard form.
(i) x – 9 + √7x3 + 6x2
(ii) √2x2 – (7 / 2)x4 + x – 5x3
(iii) 7x3 – (6 / 5)x2 + 4x – 1
(iv) y2 – √5y3 – 11 – (7 / 3) y + 9y4
Solution:
| Given Expression | Standard form of the expression |
| (i) x – 9 + √7x3 + 6x2 | √7x3 + 6x2 + x – 9 |
| (ii) √2x2 – (7 / 2)x4 + x – 5x3 | – (7 / 2)x4 – 5x3 + √2x2 + x |
| (iii) 7x3 – (6 / 5)x2 + 4x – 1 | 7x3 – (6 / 5)x2 + 4x – 1 |
| (iv) y2 – √5y3 – 11 – (7 / 3) y + 9y4 | 9y4– √5y3 + y2 – (7 / 3) y – 11 |
Question 5: Add the following polynomials and find the degree of the resultant polynomial.
(i) p (x) = 6x2 – 7x + 2 and 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:
(i) p (x) = 6x2 – 7x + 2 and q (x) = 6x3 – 7x + 15
p(x) + q(x) = 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x3 + 6x2 – 7x – 7x + 2 + 15
= 6x3 + 6x2 – 14x + 17
(ii) h (x) = 7x3 – 6x + 1, f (x) = 7x2 + 17x – 9
h(x) + f(x) = 7x3 – 6x + 1 + 7x2 + 17x – 9
= 7x3 + 7x2 – 6x + 17x + 1 – 9
= 7x3 + 7x2 + 11x – 8
(iii) f (x) = 16x4 – 5x2 + 9, g (x) = -6x3 + 7x – 15
f(x) + g(x) = 16x4 – 5x2 + 9 + -6x3 + 7x – 15
= 16x4 – 6x3 – 5x2 + 7x + 9 – 15
= 16x4 – 6x3 – 5x2 + 7x – 6
Question 6: Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x2 + 6x – 1 and q(x) = 6x – 9
(ii) f(y) = 6y2 – 7y + 2 and g(y) = 7y + y3
(iii) h(z) = z5 – 6z4 + z and f(z) = 6z2 + 10z – 7
Solution:
(i) p(x) – q(x) = (7x2 + 6x – 1) – (6x – 9)
= 7x2 + 6x – 1 – 6x + 9
= 7x2 + 6x – 6x – 1 + 9
= 7x2 + 8
Degree of the obtained polynomial is 2.
(ii) f(y) = 6y2 – 7y + 2 and g(y) = 7y + y3
f(y) – g(y) = (6y2 – 7y + 2) – (7y + y3)
= 6y2 – 7y + 2 – 7y – y3
= – y3 + 6y2 – 7y – 7y + 2
= – y3 + 6y2 – 14y + 2
Degree of the obtained polynomial is 3.
(iii) h(z) = z5 – 6z4 + z and f(z) = 6z2 + 10z – 7
h(z) – f(z) = (z5 – 6z4 + z) – (6z2 + 10z – 7)
= z5 – 6z4 + z – 6z2 – 10z + 7
= z5 – 6z4 – 6z2 – 9z + 7
Degree of the obtained polynomial is 5.
Question 7: What should be added to 2x3 + 6x2 – 5x + 8 to get 3x3 – 2x2 + 6x + 15 ?
Solution:
Let p(x) be the required polynomial to be added.
By adding p(x) and 2x3 + 6x2 – 5x + 8, we get 3x3 – 2x2 + 6x + 15
p(x) + (2x3 + 6x2 – 5x + 8) = 3x3 – 2x2 + 6x + 15
p(x) = (3x3 – 2x2 + 6x + 15) – (2x3 + 6x2 – 5x + 8)
p(x) = 3x3 – 2x3– 2x2 + 6x2 + 6x – 5x + 15 + 8
p(x) = x3 + 4x2 + x + 23
Question 8: What must be subtracted from 2x4 + 4x2 – 3x + 7 to get 3x3 – x2 + 2x + 1?
Solution:
Let p(x) be the required polynomial to be subtracted.
(2x4 + 4x2 – 3x + 7) – p(x) = 3x3 – x2 + 2x + 1
p(x) = (2x4 + 4x2 – 3x + 7) – (3x3 – x2 + 2x + 1)
= 2x4 + 4x2 – 3x + 7 – 3x3 + x2 – 2x – 1
= 2x4 – 3x3 + 4x2 + x2 – 3x – 2x + 7 – 1
p(x) = 2x4 – 3x3 + 5x2 – 5x + 6
Question 9: Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x2 – 9 and q(x) = 6x2 + 7x – 2
(ii) f(x) = 7x + 2 and g(x) = 15x – 9
(iii) h(x) = 6x2 – 7x + 1 and f(x) = 5x – 7
Solution:
(i) p(x) = x2 – 9 and q(x) = 6x2 + 7x – 2
p(x) * q(x) = (x2 – 9) * (6x2 + 7x – 2)
= (x2 * 6x2 + x2 * 7x – x2 * 2) – (9 * 6x2 + 9 * 7x – 9 * 2)
= 6x4 + 7x3 – 2x2 – 54x2 – 63x + 18
= 6x4 + 7x3 – 56x2 – 63x + 18
(ii) f(x) = 7x + 2 and g(x) = 15x – 9
f(x) * g(x) = (7x + 2) * (15x – 9)
= (7x * 15x + 7x * -9) + 2 * 15x – 2 * 9
= 105x2 – 63x + 30x – 18
= 105x2 – 33x – 18
(iii) h(x) = 6x2 – 7x + 1 and f(x) = 5x – 7
h(x) * f(x) = (6x2 – 7x + 1) * (5x – 7)
= (6x2 * 5x + 6x2 * -7) – (7x * 5x – 7 * 7) + 5x – 7
= 30x3 – 42x2 – 35x2 + 49x + 5x – 7
= 30x3 – 77x2 + 54x – 7
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 and y. If x = 10, y = 5 find the amount paid by him.
Solution:
In order to find the total amount paid by him, multiply the cost of 1 chocolate by the number of chocolates he buys.
Cost of 1 chocolate = x + y
Number of chocolates that Amir buys = x + y
Total amount = (x + y) (x + y)
= (x + y)2 —(1)
= x2 + 2xy + y2
By applying the values of x and y,
= (10 + 5)2
= 152
= 225
Hence, he has to pay Rs. 225.
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:
Length of the rectangle = 3x + 2
Breadth of the rectangle = 3x – 2
Area of rectangle = (3x + 2)(3x – 2)
= 9x2 – 6x + 6x – 4
= 9x2 – 4
If x = 20
Area of rectangle = 9(20)2 – 4
= 9(400) – 4
= 3600 – 4
= 3596
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:
The degree of the polynomial p(x) is 1.
Degree of the polynomial q(x) is 2.
The product of polynomials is 3.
Hence it is a cubic polynomial.
Samacheer Kalvi 9th Maths Chapter 3: Algebra Book Exercise 3.2 Questions and Solutions
Question 1: Find the value of the polynomial f(y) = 6y – 3y2 + 3 at
(i) y = 1
(ii) y = -1
(iii) y = 0
Solution:
f(y) = 6y – 3y2 + 3 at
(i) y = 1
f(1) = 6(1) – 3(1)2 + 3
= 6 – 3 + 3
= 6
(ii) y = -1
f(-1) = 6(-1) – 3(-1)2 + 3
= -6 – 3 + 3
= -6
(iii) y = 0
f(0) = 6(0) – 3(0)2 + 3 = 3
Question 2: If p(x) = x2 – 2√2 x + 1, find p(2√2)
Solution:
p(x) = x2 – 2√2 x + 1
p(2√2) = (2√2)2 – 2√2 (2√2) + 1
= 4(2) – 4(2) + 1
= 8 – 8 + 1
= 1
Question 3: Find the zeros of the polynomial in each of the following:
(i) p(x) = x – 3
(ii) p(x) = 2x + 5
(iii) q(y) = 2y – 3
(iv) f(z) = 8z
(v) p(x) = ax when a ≠ 0
(vi) h(x) = ax + b, a ≠ 0, a,b ∈ R
Solution:
(i) p(x) = x – 3
p(x) = 0
x – 3 = 0
x = 3
(ii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = -5
x = -5 / 2
(iii) q(y) = 2y – 3
q(y) = 0
2y – 3 = 0
2x = 3
x = 3 / 2
(iv) f(z) = 8z
f(z) = 0
8z = 0
z = 0
(iv) p(x) = ax, when a ≠ 0
p(x) = 0
ax = 0
x = 0
(vi) h(x) = ax + b, a ≠ 0, a,b ∈ R
h(x) = 0
ax + b = 0
ax = -b
x = -b / a
Question 4: Find the roots of the polynomial equations.
(i) 5x – 6 = 0
(ii) x + 3 = 0
(iii) 10x + 9 = 0
(iv) 9x – 4 = 0
Solution:
(i) 5x – 6 = 0
5x – 6 = 0
5x = 6
x = 6 / 5
(ii) x + 3 = 0
x + 3 = 0
x = -3
(iii) 10x + 9 = 0
10x + 9 = 0
10x = -9
x = -9 / 10
(iv) 9x – 4 = 0
9x – 4 = 0
9x = 4
x = 4 / 9
Question 5: Verify whether the following are zeros of the polynomial indicated against them, or not.
(i) p(x) = 2x – 1, x = ½
(ii) p(x) = x3 – 1, x = 1
(iii) p(x) = ax + b, x = -b / a
(iv) p(x) = (x + 3) (x – 4), x = 4, x = –3
Solution:
(i) p(x) = 2x – 1, x = 1/2
p(1 / 2) = 2(1 / 2) – 1
= 1 – 1
p(1 / 2) = 0
0 is obtained by applying 1/2, thus, 1 / 2 is the zero of the polynomial.
(ii) p(x) = x3 – 1, x = 1
p(x) = x3 – 1
p(1) = 13 – 1
p(1) = 0
0 is obtained by applying 1, thus, 1 is the zero of the polynomial.
(iii) p(x) = ax + b, x = -b / a
p(x) = ax + b
p(-b / a) = a(-b / a) + b
p(-b / a) = 0
0 is obtained by applying -b / a, thus, -b / a is the zero of the polynomial.
(iv) p(x) = (x + 3) (x – 4), x = 4, x = –3
p(x) = (x + 3) (x – 4)
x = 4
p(4) = (4 + 3) (4 – 4)
p(4) = 0
x = -3
p(-3) = (-3 + 3) (-3 – 4)
p(-3) = 0
Thus, 4 and -3 are the zeroes of the polynomial.
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