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

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 – 3.3

15. Solution:
p(x) = x3 – 5x2 + 4x – 3
P(2) = (2)3 – 5(2)2 + 4(2) – 3
= 8 – 5(4) + 8 – 3
= 8 – 20 + 8 – 3
= 16 – 23
= -7
p{2) ≠ 0
∴ p(x) is not a multiple of g(x)

16. (i) p(x) = x3 – 2x2 – 4x – 1; g(x) = x + 1
Solution:
p(x) = x3 – 2x2 – 4x – 1
p(-1) = (-1)3 – 2(-1)2 – 4(-1) – 1
= 1 – 2 + 4 – 1
= 4 – 4 = 0
∴ The remainder = 0

17. (ii) p(x) = 4x3 – 12x2 + 14x – 3; g(x) = 2x – 1
Solution:
p(x) = 4x3 – 12x2 + 14x – 3
= 4 × 18 – 12 × 14 + 14 × 12 – 3
= 12 – 3 + 7 – 3
= 12 – 6 + 7
= 12 + 1
= 32
∴ The reminder is 32

18. (iii) p(x) = x3 – 3x2 + 4x + 50; g(x) = x – 3
Solution:
p(x) = x3 – 3x2 + 4x + 50
p(3) = 33 – 3(3)2 + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
The remainder is 62.

19. Solution:
p(x) = 3x3 – 4x2 + 7x – 5
When it is divided by x +3,
p(-3) = 3(-3)3 – 4(-3)2 + 7(-3) – 5
= 3(-27) – 4(9) – 21 – 5
= -81 – 36 – 21 – 5
= -143
The remainder is -143.

20. Solution:
p(x) = x2018 + 2018
When it is divided by x – 1,
p(1) = 12018 + 2018
= 1 + 2018
= 2019
The remainder is 2019.

21. Solution:
p(x) = 2x3 – kx2 + 3x + 10
When it is exactly divided by x – 2,
P(2) = 0
2(2)3 – k(2)2 + 3(2) + 10 = 0
2(8) – k(4) + 6 + 10 = 0
16 – k(4) + 6 + 10 = 0
16 – 4k + 6 + 10 = 0
32 – 4k = 0
32 = 4k
∴ k = 324
= 8
The value of k = 8

22. Solution:
p(x1) = 2x3 + ax2 + 4x – 12
When it is divided by x – 3,
p(3) = 2(3)3 + a(3)2 + 4(3) – 12
= 54 + 9a + 12 – 12
= 54 + 9a ……….(R1)
p(x2) = x3 + x2 – 2x + a
When it is divided by x – 3,
p(3) = 33 + 32 – 2(3) + a
= 27 + 9 – 6 + a
= 30 + a ………(R2)
The given remainders are same (R1 = R2)
∴ 54 + 9a = 30 + a
9a – a = 30 – 54
8a = -24
∴ a = -24/8
= -3
Consider R2,
Remainder = 30 – 3
= 27

23. (i) x3 + 5x2 – 10x + 4
Solution:
p(x) = x3 + 5x2 – 10x + 4
p(1) = 13 + 5(1) – 10(1) +
4
= 1 + 5 – 10 + 4
= 10 – 10
= 0
∴ x – 1 is a factor of p(x)
(ii) x4 + 5x2 – 5x + 1
Solution:
p(1) = 14 + 5(1)2 – 5(1) + 1
= 1 + 5 – 5 + 1
= 7 – 5
= 2
= 0
∴ x – 1 is not a factor of
p(x)

24. Solution:
p(x) = 2x3 – 5x2 – 28x + 15
x – 5 is a factor
p(5) = 2(5)3 – 5(5)2 – 28(5) + 15
= 250 – 125 – 140 + 15
= 265 – 265
= 0
∴ x – 5 is a factor of p(x)

25. Solution:
p(x) = x3 – 3x2 – mx + 24
when x + 3 is a factor
P(-3) = 0
(-3)3 – 3(-3)2 – m(-3) + 24 =
0
-27 – 27 + 3m + 24 = 0
-54 + 24 + 3m = 0
-30 + 3m = 0
3m = 30
m = 303
= 10
The value of m = 10

26. Solution:
p(x) = ax2 + 5x + b
when (x-2) is a factor
P(2) = 0
a(2)2 + 5(2) + b = 0
4a + 10 + b = 0
4a + b = -10 …….(1)
when (x – 12) is a factor
p(12) = 0
a(12)2 + 5(12) + b = 0
Multiply by 4
a + 10 + 4b = 0
a + 46 = -10 …….(2)
From (1) and (2) we get
4a + b = a + 4b
4a – a = 4b – b
3a = 3b
a = b
Hence it is proved.

27. Solution:
p(x) = kx3 – 2x2 + 25x – 26
When it is divided by x – 1
P(1) = 0
k(1)3 – 2(1)2 + 25(1) – 26 = 0
k – 2 + 25 – 26 = 0
k + 25 – 28 = 0
k – 3 = 0
k = 3
The value of k = 3

28. Solution:
Let the area of a rectangle be p(x)
p(x) = x2 – 2x – 8
When x + 2 is the side of the
rectangle
p(-2) = (-2)2 – 2(-2) – 8
= 4 + 4 – 8
= 8 – 8
= 0
When x – 4 is the side of the rectangle.
P(4) = (4)2 – 2(4) – 8
= 16 – 8 – 8
= 16 – 16
= 0
(x + 2) and (x – 4) are the sides of a
rectangle