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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 x^{2} is not a whole number, but it is (\(\frac{1}{x^2}\) = x^{-2}) negative number.

(ii) x^{2} (x – 1)

Solution:

x^{2} (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. (\(\frac{1}{x^{-2}}\) = x^{2} and \(\frac{1}{x^{-1}}\) = x)

(v) √5x^{2} + √3x + √2

Solution:

√5x^{2} + √3x + √2 is a polynomial.

(vi) m^{2} – \(\sqrt[3]{m}\) + 7m – 10

m^{2} –\(\sqrt[3]{m}\) + 7m – 10 is not a polynomial. Since the exponent of m is not a whole number.

(\(\sqrt[3]{m}\) = m^{1/3})

Question 2.

Write the coefficient of x^{2} and x in each of the following polynomials.

(i) 4 + \(\frac{2}{5}\) x^{2} – 3x

Solution:

Coefficient of x^{2} is \(\frac{2}{5}\) and coefficient of x is -3.

(ii) 6 – 2x^{2} + 3x^{3} – √7x

Solution:

Coefficient of x^{2 }is -2 and coefficient of x is -√7

(iii) π x^{2} – x + 2

Solution:

Coefficient of x^{2} is π and coefficient of x is -1.

(iv) √3x^{2} + √2x + 0.5

Solution:

Coefficient of x^{2} is √3 and coefficient of x is √2

(v) x^{2} – \(\frac{7}{2}\) x + 8

Solution:

Coefficient of x^{2} is 1 and coefficient of x is –\(\frac{7}{2}\)

Question 3.

Find the degree of the following polynomials.

(i) 1 – √2 y^{2} + y^{7}

(ii) \(\frac{x^{3}-x^{4}+6 x^{6}}{x^{2}}\)

(iii) x^{3} (x^{2} + x)

(iv) 3x^{4} + 9x^{2} + 27x^{6}

(v) 2√5p^{4} \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)

Solution:

(i) 1 – √2 y^{2} + y^{7}

The degree of the polynomial is 7.

(ii)

= x – x^{2} + 6x^{4}

The degree of the polynomial is 4.

(iii) x^{3} (x^{2} + x) = x^{5} + x^{4}

The degree of the polynomial is 5.

(iv) 3x^{4} + 9x^{2} + 27x^{6}

The degree of the polynomial is 6.

(v) 2√5p^{4} \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)

The degree of the polynomial is 4.

Question 4.

Rewrite the following polynomial in standard form.

(i) x – 9 + √7x^{3} + 6x^{2}

Solution:

The standard form is √7x^{3} + 6x^{2} – x – 9

(or) – 9 + x + 6x^{2} + √7x^{3}

(ii) √2x^{2} – \(\frac{7}{2}\) x^{4} + x – 5x^{3}

Solution:

The standard form is – \(\frac{7}{2}\) x^{4} – 5x^{3} + √2x^{2} + x

(or) x + √2x^{2} – 5x^{3} – \(\frac{7}{2}\) x^{4}

(iii) 7x^{3} – \(\frac{6}{5}\) x^{2} + 4x – 1

Solution:

The given polynomial is in standard form (or) – 1 + 4x – \(\frac{6}{5}\) x^{2} + 7x^{3}

(iv) y^{2} + √5y^{3} – 11 – \(\frac{7}{3}\) y + 9y^{4}

Solution:

The standard form is 9y^{4} + √5y^{3} + y^{2} – \(\frac{7}{3}\) y – 11

(or) – 11 – \(\frac{7}{3}\) y + y^{2} + √5y^{3} + 9y^{4}

Question 5.

Add the following polynomials and find the degree of the resultant polynomial

(i) p(x) = 6x^{2} – 7x + 2; q(x) = 6x^{3} – 7x + 15

Solution:

p(x) + q(x) = 6x^{2} – 7x + 2 + 6x^{3} – 7x + 15

= 6x^{3} + 6x^{2} – 7x – 7x + 2 + 15

= 6x^{3} + 6x^{2} – 14x + 17

The degree of the polynomial is 3.

(ii) h(x) = 7x^{3} – 6x + 1; f(x) = 7x^{2} + 17x – 9

Solution:

h(x) + f(x) = 7x^{3} – 6x + 1 + 7x^{2} + 17x – 9

= 7x^{3} + 7x^{2} + 11x – 8

The degree of the polynomial is 3.

(iii) f(x) = 16x^{4} – 5x^{2} + 9; g(x) = -6x^{3} + 7x – 15

Solution:

f(x) + g(x) = 16x^{4} – 5x^{2} + 9 – 6x^{3} + 7x – 15

= 16x^{4} – 6x^{3} – 5x^{2} + 7x + 9 – 15

= 16x^{4} – 6x^{3} – 5x^{2} + 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) = 7x^{2} + 6x – 1; q(x) = 6x – 9

Solution:

p(x) – q(x) = 7x^{2} + 6x – 1 – (6x – 9)

= 7x^{2} + 6x – 1 – 6x + 9

= 7x^{2} + 6x – 6x – 1 + 9

= 7x^{2} + 8

The degree of the polynomial is 2.

(ii) f(y) = 6y^{2} – 7y + 2; g(y) = 7y + y^{3}

Solution:

f(y) – g(y) = 6y^{2} – 7y + 2 – (7y + y^{3})

= 6y^{2} – 7y + 2 – 7y – y^{3}

= -y^{3} + 6y^{2} – 7y – 7y + 2

= -y^{3} + 6y^{2} – 14y + 2

The degree of the polynomial is 3.

(iii) h(z) = z^{5} – 6z^{4} + z; f(z) = 6z^{2} + 10z – 7

Solution:

h(z) – f(z) = z^{5} – 6z^{4} + z – (6z^{2} + 10z – 7)

= z^{5} – 6z^{4} + z – 6z^{2} – 10z + 7

= z^{5} – 6z^{4} – 6z^{2} + z – 10z + 7

= z^{5} – 6z^{4} – 6z^{2} – 9z + 7

The degree of the polynomial is 5.

Question 7.

What should be added to 2x^{3} + 6x^{2} – 5x + 8 to get 3x^{3} – 2x^{2} + 6x + 15?

Solution:

3x³ – 2x^{2} + 6x + 15 – (2x³ + 6x^{2} – 5x + 8)

= 3x³ – 2x^{2} + 6x + 15 – 2x³ – 6x^{2} + 5x – 8

= 3x³ – 2x³- 2x^{2} – 6x^{2} + 6x + 5x + 15 – 8

= x³ – 8x^{2} + 11x + 7

x³ – 8x^{2} + 11x + 7 must be added to get 3x³ – 2x^{2 }+ 6x + 15.

Question 8.

What must be subtracted from 2x^{4} + 4x^{2} – 3x + 7 to get 3x^{3} – x^{2} + 2x + 1?

Solution:

2x^{4} + 4x^{2} – 3x + 7 – (3x^{3} – x^{2} + 2x + 1)

= 2x^{4} + 4x^{2} – 3x + 7 – 3x^{3} + x^{2} – 2x – 1

= 2x^{4} – 3x^{3} + 4x^{2} + x^{2} – 3x – 2x + 7 – 1

= 2x^{4} – 3x^{3} + 5x^{2} – 5x + 6

2x^{4} – 3x^{3} + 5x^{2} – 5x + 6 must be subtracted to get 3x^{3} – x^{2} + 2x + 1.

Question 9.

Multiply the following polynomials and find the degree of the resultant polynomial:

(i) p(x) = x^{2} – 9, q(x) = 6x^{2} + 7x – 2

Solution:

p(x) × q(x) = (x^{2} – 9) (6x^{2} + 7x – 2)

= 6x^{4} + 7x^{3} – 2x^{2} – 54x^{2} – 63x + 18

= 6x^{4} + 7x^{3} – 56x^{2 }– 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)

= 105x^{2} – 63x + 30x – 18

= 105x^{2} – 33x – 18

The degree of the polynomial is 2.

(iii) h(x) = 6x^{2} – 7x + 1, f(x) = 5x – 7

Solution:

h(x) × f(x) = (6x^{2} – 7x + 1) (5x – 7)

= 30x^{3} – 42x^{2} – 35x^{2} + 49x + 5x – 7

= 30x^{3} – 77x^{2} + 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)

= x^{2} + xy + xy + y^{2}

= x^{2} + 2xy + y^{2}

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)

= 9x^{2} – 6x + 6x – 4

= 9x^{2} – 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.