Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Polynomials, Constants, variables, algebraic expression

1. PEDAGOGY OF
MATHEMATICS – PART II
By
Dr. I. Uma Maheswari
Principal
Peniel Rural College of Education,Vemparali, Dindigul District
iuma_maheswari@yahoo.co.in

2. STD IX
CHAPTER 3 – ALGEBRA
Ex – 3.1

35. Given Expression Is it a
polynomial?
Reason
(i) (1 / x
2
) + 3x – 4 = x
-2
+ 3x – 4 No Negative integral power
(ii) x
2
(x – 1) Yes Positive integral power
(iii) (1 / x) (x + 5) = x
-1
* (x + 5)
= x
-1+1
* 5x
-1
= x
0
+ 5x
-1
No One of the integral powers is
negative.
(iv) (1 / x
-2
) + ( 1 / x
-1
) + 7 = x
2
+ x + 7 Yes Positive integral power
(v) √5 x
2
+ √3 x + √2 Yes Positive integral power
(vi) m
2
– ∛m + 7m – 10
= m
2
– m
1/3
+ 7m – 10
No One of the powers is fractional.
Solution:

37. Given Expression Coefficient of x
2
Coefficient of x
(i) 4 + (2 / 5)x
2
– 3x 2 / 5 -3
(ii) 6 – 2x
2
+ 3x
3
– √7 x -2 – √7
(iii) πx
2
– x + 2 π -1
(iv) √3 x
2
+ √2x + 0.5 √3 √2
(v) x
2
– (7 / 2)x + 8 1 – (7 / 2)
Solution:

39. Given Expression Degree of the polynomial
(i) 1 – √2y
2
+ y
7
7
(ii) (x
3
– x
4
+ 6x
6
) / x
2
4
(iii) x
3
(x
2
+ x) 5
(iv) 3x
4
+ 9x
2
+ 27x
6
6
(v) 2√5p
4
– (8p
3
/ √3) + (2p
2
/ 7) 4
Solution:

41. Given Expression Standard form of the expression
(i) x – 9 + √7x
3
+ 6x
2
√7x
3
+ 6x
2
+ x – 9
(ii) √2x
2
– (7 / 2)x
4
+ x – 5x
3
– (7 / 2)x
4
– 5x
3
+ √2x
2
+ x
(iii) 7x
3
– (6 / 5)x
2
+ 4x – 1 7x
3
– (6 / 5)x
2
+ 4x – 1
(iv) y
2
– √5y
3
– 11 – (7 / 3) y + 9y
4
9y
4
– √5y
3
+ y
2
– (7 / 3) y – 11
Solution:

42. Solution:
(i) p (x) = 6x2 – 7x + 2 and q (x) = 6x3 – 7x + 15
p(x) + q(x) = 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x2 – 7x + 2 + 6x3 – 7x + 15
= 6x3 + 6x2 – 7x – 7x + 2 + 15
= 6x3 + 6x2 – 14x + 17

43. (ii) h (x) = 7x3 – 6x + 1, f (x) = 7x2 + 17x – 9
h(x) + f(x) = 7x3 – 6x + 1 + 7x2 + 17x – 9
= 7x3 + 7x2 – 6x + 17x + 1 – 9
= 7x3 + 7x2 + 11x – 8
(iii) f (x) = 16x4 – 5x2 + 9, g (x) = -6x3 + 7x – 15
f(x) + g(x) = 16x4 – 5x2 + 9 + -6x3 + 7x – 15
= 16x4 – 6x3 – 5x2 + 7x + 9 – 15
= 16x4 – 6x3 – 5x2 + 7x – 6

44. Solution:
(i) p(x) – q(x) = (7x2 + 6x – 1) – (6x – 9)
= 7x2 + 6x – 1 – 6x + 9
= 7x2 + 6x – 6x – 1 + 9
= 7x2 + 8
Degree of the obtained polynomial is 2.

45. (ii) f(y) = 6y2 – 7y + 2 and g(y) = 7y + y3
f(y) – g(y) = (6y2 – 7y + 2) – (7y + y3)
= 6y2 – 7y + 2 – 7y – y3
= – y3 + 6y2 – 7y – 7y + 2
= – y3 + 6y2 – 14y + 2
Degree of the obtained polynomial is 3.
(iii) h(z) = z5 – 6z4 + z and f(z) = 6z2 + 10z – 7
h(z) – f(z) = (z5 – 6z4 + z) – (6z2 + 10z – 7)
= z5 – 6z4 + z – 6z2 – 10z + 7
= z5 – 6z4 – 6z2 – 9z + 7
Degree of the obtained polynomial is 5.

46. Solution:
Let p(x) be the required polynomial to be added.
By adding p(x) and 2x3 + 6x2 – 5x + 8, we get 3x3 – 2x2 + 6x + 15
p(x) + (2x3 + 6x2 – 5x + 8) = 3x3 – 2x2 + 6x + 15
p(x) = (3x3 – 2x2 + 6x + 15) – (2x3 + 6x2 – 5x + 8)
p(x) = 3x3 – 2x3– 2x2 + 6x2 + 6x – 5x + 15 + 8
p(x) = x3 + 4x2 + x + 23

47. Solution:
Let p(x) be the required polynomial to be subtracted.
(2x4 + 4x2 – 3x + 7) – p(x) = 3x3 – x2 + 2x + 1
p(x) = (2x4 + 4x2 – 3x + 7) – (3x3 – x2 + 2x + 1)
= 2x4 + 4x2 – 3x + 7 – 3x3 + x2 – 2x – 1
= 2x4 – 3x3 + 4x2 + x2 – 3x – 2x + 7 – 1
p(x) = 2x4 – 3x3 + 5x2 – 5x + 6

48. Solution:
(i) p(x) = x2 – 9 and q(x) = 6x2 + 7x – 2
p(x) * q(x) = (x2 – 9) * (6x2 + 7x – 2)
= (x2 * 6x2 + x2 * 7x – x2 * 2) – (9 * 6x2 + 9 * 7x – 9 * 2)
= 6x4 + 7x3 – 2x2 – 54x2 – 63x + 18
= 6x4 + 7x3 – 56x2 – 63x + 18

49. (ii) f(x) = 7x + 2 and g(x) = 15x – 9
f(x) * g(x) = (7x + 2) * (15x – 9)
= (7x * 15x + 7x * -9) + 2 * 15x – 2 * 9
= 105x2 – 63x + 30x – 18
= 105x2 – 33x – 18
(iii) h(x) = 6x2 – 7x + 1 and f(x) = 5x – 7
h(x) * f(x) = (6x2 – 7x + 1) * (5x – 7)
= (6x2 * 5x + 6x2 * -7) – (7x * 5x – 7 * 7) + 5x – 7
= 30x3 – 42x2 – 35x2 + 49x + 5x – 7
= 30x3 – 77x2 + 54x – 7

50. Solution:
In order to find the total amount paid by him, multiply the cost of 1 chocolate by the number of
chocolates he buys.
Cost of 1 chocolate = x + y
Number of chocolates that Amir buys = x + y
Total amount = (x + y) (x + y)
= (x + y)2 —(1)
= x2 + 2xy + y2
By applying the values of x and y,
= (10 + 5)2
= 152
= 225
Hence, he has to pay Rs. 225.

51. Solution:
Length of the rectangle = 3x + 2
Breadth of the rectangle = 3x – 2
Area of rectangle = (3x + 2)(3x – 2)
= 9x2 – 6x + 6x – 4
= 9x2 – 4
If x = 20
Area of rectangle = 9(20)2 – 4
= 9(400) – 4
= 3600 – 4
= 3596

52. Solution:
The degree of the polynomial p(x) is 1.
Degree of the polynomial q(x) is 2.
The product of polynomials is 3.
Hence it is a cubic polynomial.