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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: ∴ The quotient = 4x2 – 6x – 5
The remainder = 33

(ii) (8y3 – 16y2 + 16y – 15) ÷ (2y – 1)
Solution: ∴ The quotient = 4y2 – 6y + 5
The remainder = -10 (iii) (8x3 – 1) ÷ (2x – 1)
Solution: ∴ The quotient = 4x2 + 2x + 1
The remainder = 0

(iv) (-18z + 14z2 + 24z3 + 18) ÷ (3z + 4)
Solution: ∴The quotient = 8z2 – 6z + 2
The remainder = 10 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) Length of the rectangle = x + 4  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 Height of the parallelogram = 5x – 4  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 Mean = x2 – 5x + 25 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] x – 3 = 0
x = 3
Hence the quotient = x2 + 4x + 5
Remainder = 12 (ii) (x3 + 2x2 – x – 4) ÷ (x + 2)
Solution:
p(x) = x3 + 2x2 -x – 4
d(x) = x + 2 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 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  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 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 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 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