Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Question 1.
Find the quotient and remainder of the following.
(i) 4x3 + 6x2 – 23x + 18) ÷ (x + 3)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 1
∴ The quotient = 4x2 – 6x – 5
The remainder = 33

(ii) (8y3 – 16y2 + 16y – 15) ÷ (2y – 1)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 2
∴ The quotient = 4y2 – 6y + 5
The remainder = -10

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

(iii) (8x3 – 1) ÷ (2x – 1)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 3
∴ The quotient = 4x2 + 2x + 1
The remainder = 0

(iv) (-18z + 14z2 + 24z3 + 18) ÷ (3z + 4)
Solution:
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 4
∴The quotient = 8z2 – 6z + 2
The remainder = 10

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Question 2.
The area of a rectangle is x2 + 7x + 12. If its breadth is (x + 3) then find its length.
Solution:
Let the length of the rectangle be “l”
The breadth of the rectangle = x + 3
Area of the rectangle = length × breadth
x2 + 7x + 12 = l(x + 3)
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 5
Length of the rectangle = x + 4
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 6

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Question 3.
The base of a parallelogram is (5x + 4). Find its height if the area is 25x2 – 16.
Solution:
Let the height of the parallelogram be “h”.
Base of the parallelogram = 5x + 4
Area of a parallelogram = 25x2 – 16
∴ Base x Height = 25x2 – 16
(5x + 4) × h = 25x2– 16
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 7
Height of the parallelogram = 5x – 4
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 8

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Question 4.
The sum of (x + 5) observations is (x3 + 125). Find the mean of the observations.
Solution:
Sum of the observation = x3 + 125
Number of observation = x + 5
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 9
Mean = x2 – 5x + 25
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 10

Question 5.
Find the quotient and remainder for the following using synthetic division:
(i) (x3 + x2 – 7x – 3) ÷ (x – 3)
Solution:
p(x) = x3 + x2 – 7x – 3
d(x) = x – 3 [p(x) = d(x) × q(x) + r]
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 11
x – 3 = 0
x = 3
Hence the quotient = x2 + 4x + 5
Remainder = 12

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

(ii) (x3 + 2x2 – x – 4) ÷ (x + 2)
Solution:
p(x) = x3 + 2x2 -x – 4
d(x) = x + 2
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 12
x + 2 = 0
x = -2
The quotient = x2 – 1
Remainder = -2

(iii) (3x3 – 2x2 + 7x – 5) ÷ (x + 3)
Solution:
p(x) = 3x3 – 2x2 + 7x – 5
d(x) = x + 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 13
x + 3 = 0
x = -3
The quotient = 3x2 – 11x + 40
Remainder = -125

(iv) (8x4 – 2x2 + 6x + 5) ÷ (4x + 1)
Solution:
p(x) = 8x4 – 2x2 + 6x + 5
d(x) = 4x + 1
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 14

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Question 6.
If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x + 1) is (4x3 + px2 -qx + 3), then find p, q and also the remainder.
Solution:
p(x) = 8x4 – 2x2 + 6x – 7
d(x) = 2x + 1
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 15
2x + 1 = 0
2x = -1
x = –\(\frac{1}{2}\)
The quotient = \(\frac{1}{2}\) [8x3 – 4x2 + 6]
= 4x3 – 2x2 + 3
= 4x3 – 2x2 + 0x + 3
The given quotient is = 4x3 + px2 – qx + 3
(compared with the given quotient)
The value of p = -2 and q = 0
Remainder = -10

Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7

Question 7.
If the quotient obtained on dividing 3x3 + 11x2 + 34x + 106 by x – 3 is 3x2 + ax + b, then find a, b and also the remainder.
Solution:
p(x) = 3x3 + 11x2 + 34x + 106
d(x) = x – 3
Samacheer Kalvi 9th Maths Guide Chapter 3 Algebra Ex 3.7 16
x – 3 = 0
x = 3
The quotient is = 3x2 + 20x + 94
The given quotient is = 3x2 + ax + b
Compared with the given quotient
The value of a = 20 and b = 94
The remainder = 388

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