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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1
Question 1.
Which of the following expressions are polynomials. If not give reason:
(i) \(\frac{1}{x^2}\) + 3x – 4
Solution:
(i) \(\frac{1}{x^2}\) + 3x – 4 is not a polynomial. Since the exponent of x2 is not a whole number, but it is (\(\frac{1}{x^2}\) = x-2) negative number.
(ii) x2 (x – 1)
Solution:
x2 (x – 1) is a polynomial.
(iii) \(\frac{1}{x}\) (x + 5)
Solution:
\(\frac{1}{x}\) (x + 5) is not a polynomial. Since the exponent of x is not a whole number, but it is (\(\frac{1}{x}\) = x-1) negative number.
(iv) \(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7
Solution:
\(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7 is a polynomial. (\(\frac{1}{x^{-2}}\) = x2 and \(\frac{1}{x^{-1}}\) = x)
(v) √5x2 + √3x + √2
Solution:
√5x2 + √3x + √2 is a polynomial.
(vi) m2 – \(\sqrt[3]{m}\) + 7m – 10
m2 –\(\sqrt[3]{m}\) + 7m – 10 is not a polynomial. Since the exponent of m is not a whole number.
(\(\sqrt[3]{m}\) = m1/3)
Question 2.
Write the coefficient of x2 and x in each of the following polynomials.
(i) 4 + \(\frac{2}{5}\) x2 – 3x
Solution:
Coefficient of x2 is \(\frac{2}{5}\) and coefficient of x is -3.
(ii) 6 – 2x2 + 3x3 – √7x
Solution:
Coefficient of x2 is -2 and coefficient of x is -√7
(iii) π x2 – x + 2
Solution:
Coefficient of x2 is π and coefficient of x is -1.
(iv) √3x2 + √2x + 0.5
Solution:
Coefficient of x2 is √3 and coefficient of x is √2
(v) x2 – \(\frac{7}{2}\) x + 8
Solution:
Coefficient of x2 is 1 and coefficient of x is –\(\frac{7}{2}\)
Question 3.
Find the degree of the following polynomials.
(i) 1 – √2 y2 + y7
(ii) \(\frac{x^{3}-x^{4}+6 x^{6}}{x^{2}}\)
(iii) x3 (x2 + x)
(iv) 3x4 + 9x2 + 27x6
(v) 2√5p4 \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
Solution:
(i) 1 – √2 y2 + y7
The degree of the polynomial is 7.
(ii) 
= x – x2 + 6x4
The degree of the polynomial is 4.
(iii) x3 (x2 + x) = x5 + x4
The degree of the polynomial is 5.
(iv) 3x4 + 9x2 + 27x6
The degree of the polynomial is 6.
(v) 2√5p4 \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
The degree of the polynomial is 4.
Question 4.
Rewrite the following polynomial in standard form.
(i) x – 9 + √7x3 + 6x2
Solution:
The standard form is √7x3 + 6x2 – x – 9
(or) – 9 + x + 6x2 + √7x3
(ii) √2x2 – \(\frac{7}{2}\) x4 + x – 5x3
Solution:
The standard form is – \(\frac{7}{2}\) x4 – 5x3 + √2x2 + x
(or) x + √2x2 – 5x3 – \(\frac{7}{2}\) x4
(iii) 7x3 – \(\frac{6}{5}\) x2 + 4x – 1
Solution:
The given polynomial is in standard form (or) – 1 + 4x – \(\frac{6}{5}\) x2 + 7x3
(iv) y2 + √5y3 – 11 – \(\frac{7}{3}\) y + 9y4
Solution:
The standard form is 9y4 + √5y3 + y2 – \(\frac{7}{3}\) y – 11
(or) – 11 – \(\frac{7}{3}\) y + y2 + √5y3 + 9y4
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
Solution:
p(x) + q(x) = 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x3 + 6x2 – 7x – 7x + 2 + 15
= 6x3 + 6x2 – 14x + 17
The degree of the polynomial is 3.
(ii) h(x) = 7x3 – 6x + 1; f(x) = 7x2 + 17x – 9
Solution:
h(x) + f(x) = 7x3 – 6x + 1 + 7x2 + 17x – 9
= 7x3 + 7x2 + 11x – 8
The degree of the polynomial is 3.
(iii) f(x) = 16x4 – 5x2 + 9; g(x) = -6x3 + 7x – 15
Solution:
f(x) + g(x) = 16x4 – 5x2 + 9 – 6x3 + 7x – 15
= 16x4 – 6x3 – 5x2 + 7x + 9 – 15
= 16x4 – 6x3 – 5x2 + 7x – 6
The degree of the polynomial is 4.
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
Solution:
p(x) – q(x) = 7x2 + 6x – 1 – (6x – 9)
= 7x2 + 6x – 1 – 6x + 9
= 7x2 + 6x – 6x – 1 + 9
= 7x2 + 8
The degree of the polynomial is 2.
(ii) f(y) = 6y2 – 7y + 2; g(y) = 7y + y3
Solution:
f(y) – g(y) = 6y2 – 7y + 2 – (7y + y3)
= 6y2 – 7y + 2 – 7y – y3
= -y3 + 6y2 – 7y – 7y + 2
= -y3 + 6y2 – 14y + 2
The degree of the polynomial is 3.
(iii) h(z) = z5 – 6z4 + z; f(z) = 6z2 + 10z – 7
Solution:
h(z) – f(z) = z5 – 6z4 + z – (6z2 + 10z – 7)
= z5 – 6z4 + z – 6z2 – 10z + 7
= z5 – 6z4 – 6z2 + z – 10z + 7
= z5 – 6z4 – 6z2 – 9z + 7
The degree of the polynomial is 5.
Question 7.
What should be added to 2x3 + 6x2 – 5x + 8 to get 3x3 – 2x2 + 6x + 15?
Solution:
3x³ – 2x2 + 6x + 15 – (2x³ + 6x2 – 5x + 8)
= 3x³ – 2x2 + 6x + 15 – 2x³ – 6x2 + 5x – 8
= 3x³ – 2x³- 2x2 – 6x2 + 6x + 5x + 15 – 8
= x³ – 8x2 + 11x + 7
x³ – 8x2 + 11x + 7 must be added to get 3x³ – 2x2 + 6x + 15.
Question 8.
What must be subtracted from 2x4 + 4x2 – 3x + 7 to get 3x3 – x2 + 2x + 1?
Solution:
2x4 + 4x2 – 3x + 7 – (3x3 – x2 + 2x + 1)
= 2x4 + 4x2 – 3x + 7 – 3x3 + x2 – 2x – 1
= 2x4 – 3x3 + 4x2 + x2 – 3x – 2x + 7 – 1
= 2x4 – 3x3 + 5x2 – 5x + 6
2x4 – 3x3 + 5x2 – 5x + 6 must be subtracted to get 3x3 – x2 + 2x + 1.
Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x2 – 9, q(x) = 6x2 + 7x – 2
Solution:
p(x) × q(x) = (x2 – 9) (6x2 + 7x – 2)
= 6x4 + 7x3 – 2x2 – 54x2 – 63x + 18
= 6x4 + 7x3 – 56x2 – 63x + 18
The degree of the polynomial is 4.
(ii) f(x) = 7x + 2, g(x) = 15x – 9
Solution:
f(x) × g(x) = (7x + 2) (15x – 9)
= 105x2 – 63x + 30x – 18
= 105x2 – 33x – 18
The degree of the polynomial is 2.
(iii) h(x) = 6x2 – 7x + 1, f(x) = 5x – 7
Solution:
h(x) × f(x) = (6x2 – 7x + 1) (5x – 7)
= 30x3 – 42x2 – 35x2 + 49x + 5x – 7
= 30x3 – 77x2 + 54x – 7
The degree of the polynomial is 3.
Question 10.
The cost of a 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:
The cost of a chocolate = (x + y)
Number of chocolates bought by Amir = x + y
Total amount paid by him = (x + y) (x + y)
= x2 + xy + xy + y2
= x2 + 2xy + y2
When x = 10 and y = 5
The total amount paid by him = (10)2 + 2(10)(5) + (5)2
= 100 + 100 + 25 = 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 units
Breadth of the rectangle = 3x – 2 units
Area of the rectangle = (3x + 2) (3x – 2)
= 9x2 – 6x + 6x – 4
= 9x2 – 4
When x = 20
Area of the rectangle = 9(20)2 – 4
= 9(400) – 4
= 3600 – 4
= 3596 sq.units.
Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial is p(x) × q(x)?
Solution:
Degree of the polynomial p(x) = 1
Degree of the polynomial q(x) = 2
Degree of p(x) × q(x) = 3
The polynomial is a cubic polynomial (or) Polynomial of degree 3.