You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.7

**9th Maths Algebra Exercise 3.7 Question 1.**

Find the quotient and remainder of the following.

(i) (4x^{3} + 6x^{2} – 23x + 18) ÷ (x + 3)

(ii) (8y^{3} – 16y^{2} + 16y -15) ÷ (2y – 1)

(iii) (8x^{3} – 1) ÷ (2x – 1)

(iv) (-18z + 14z^{2} + 24z^{3} + 18) ÷ (3z + 4)

Solution:

(i) (4x^{3} + 6x^{2} – 23x + 18) ÷ (x + 3)

Quotient = 4x^{2} – 6x – 5

Remainder = 33

(ii) (8y^{3} – 16y^{2} + 16y -15) ÷ (2y – 1)

Quotient = 4y^{2} – 6y + 5

Remainder = -10

(iii) (8x^{3} – 1) ÷ (2x – 1)

Quotient = 4x^{2} + 2x + 1

Remainder = 0

(iv) (-18z + 14z^{2} + 24z^{3} + 18) ÷ (3z + 4)

Quotient = 8z^{2} – 6z + 2

Remainder = 10

**9th Maths Exercise 3.7 Question 2.**

The area of rectangle is x^{2} + 7x + 12. If its breadth is (x + 3), then find its length.

Solution:

Area of a rectangle = x^{2} + 7x + 12

Its breadth = x + 3

Area = breadth × length

x^{2} + 7x + 12 = (x + 3) × length

∴ Length = (x^{2} + 7x + 12) ÷ (x + 3)

Quotient = x + 4; Remainder = 0;

∴ Length = x + 4

**Exercise 3.7 Class 9 Question 3.**

The base of a parallelogram is (5x + 4). Find its height, if the area is 25x^{2} – 16.

Solution:

The base of a parallelogram is (5x + 4)

Area = 25x^{2} – 16

Area = b × h = 25x^{2} – 16

base = 5x + 4

∴ height of the parallelogram = 5x – 4.

**9th Standard Maths Exercise 3.7 Question 4.**

The sum of (x + 5) observations is (x^{3} + 125). Find the mean of the observations.

Solution:

The sum of (x + 5) observations is (x^{3} + 125)

Quotient = x^{2} – 5x + 25

Remainder = 0

∴ Mean of the observations = x^{2} – 5x + 25

**9th Maths 3.7 Question 5.**

Find the quotient and remainder for the following using synthetic division:

(i) (x^{3} + x^{2} – 7x – 3) ÷ (x – 3)

(ii) (x^{3} + 2x^{2} – x – 4) ÷ (x + 2)

(iii) (3x^{3} – 2x^{2} + 7x – 5) ÷ (x + 3)

(iv) (8x^{4} – 2x^{2} + 6x + 5) ÷ (4x + 1)

Solution:

(i) (x^{3} + x^{2} – 7x – 3) ÷ (x – 3)

Let p (x) = x^{3} + x^{2} – 7x – 3

q (x) = x – 3 To find the zero of x – 3

p(x) in standard form ((i.e.) descending order)

x^{3} + x^{2} – 7x – 3

Quotient is x^{2} + 4x + 5

Remainder is 12

(ii) (x^{3} + 2x^{2} – x – 4) + (x + 2)

p (x) = x^{3} + 2x^{2} – x – 4

Co-efficients are 1 2 -1 – 4

To find zero of x + 2, put x + 2 = 0; x = -2

∴ Quotient is x^{2} – 1

Remainder is -2

(iii) (3x^{3} – 2x^{2} + 7x – 5) ÷ (x + 3)

To find zero of the divisor (x + 3), put x + 3 = 0; ∴ x = – 3

Dividend in Standard form 3x^{3} – 2x^{2} + 7x – 5

Co-efficients are 3 -2 7 – 5

Synthetic Division

Quotient is 3x^{2} – 11x + 40

Remainder is -125

(iv) (8x^{4} – 2x^{2} + 6x + 5) ÷ (4x + 1)

To find zero of the divisor 4x + 1, put 4x + 1 = 0 ; 4x = -1; x = \(-\frac{1}{4}\)

Dividend in Standard form 8x^{4} + 0x^{3} – 2x^{2} + 6x + 5

Co-efficients are 8 0 -2 6 5

Synthetic Division

**Samacheer Kalvi 9th Maths Chapter 3 Question 6.**

If the quotient obtained on dividing (8x^{4} – 2x^{2} + 6x – 7) by (2x + 1) is (4x^{3} + px^{2} -qx + 3), then find p, q and also the remainder.

Solution:

Let p (x) = 8x^{4} – 2x^{2} + 6x – 7

Standard form = 8x^{4} + 0x^{3} – 2x^{2} + 6x – 7

Quotient 4x^{3} – 2x^{2} + 3 is compared with the given quotient 4x^{3} + px^{2} – qx + 3

Co-efficients of x^{2} is p = – 2

Co-efficients of x is q = 0

Remainder is – 10

**Samacheer Kalvi 9th Maths Book Answers Question 7.**

If the quotient obtained on dividing 3x^{3} + 11x^{2} + 34x + 106 by x – 3 is 3x^{2} + ax + b, then find a, b and also the remainder.

Solution:

Let p (x) = 3x^{3} + 11x^{2} + 34x + 106

p (x) in standard form

Co-efficients are 3 11 34 106

q(x) = x – 3, its zero x = 3

Synthetic division

Quotient is 3x^{2} + 20x + 94, it is compared with the given quotient 3x^{2} + ax + b

Co-efficient of x is a = 20

Constant term is b = 94

Remainder r = 388