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 – 3.5

12. Solution:
(i) 2a² + 4a²b + 8a²c = 2a²(1 + 2b + 4c)
(ii) ab – ac – mb + mc = a(b – c) – m(b – c)
= (b – c) (a – m)

13. Solution:
(i) x2 + 4x + 4 = x2 + 2 × x × 2 + 22 [a2 + 2ab+ b2 = (a + b)2]
= (x + 2)2
(ii) 3a2 – 24ab + 48b2 = 3[a2 – 8ab + 16b2]
= 3[a2 – 2 × a × 4b + (4b)2]
= 3(a- 4b)2 [a2 – 2ab + b2 = (a – b)2]

16. Solution:
[a2 + b2 + c2 + 2ab + 2bc + 2ac = (a + b + c)2]
= (2x)2 + (3y)2 + (5z)2 + 2(2x) (3y) + 2(3y) (5z) + 2(5z) (2x)
= (2x + 3y + 5z)2
(ii) 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30xz
Solution:
= (5x)2 + (2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (- 3z) + 2(-3z)
(5x)
2

17. Solution:
(i) 8x3 + 125y3 = (2x)3 + (5y)3 [a3 + b3 = (a + b)(a2 – ab + b2)
= (2x + 5y) [(2x)2 – (2x) (5y) + (5y)2]
= (2x + 5y) (4x2 – 10xy + 25y2)
(ii) 27x3 – 8y3 = (3x)3 – (2y)3 [a3 – b3 = (a – b)(a2 + ab + b2)]
= (3x – 2y) [(3x)2 + (3x) (2y) + (2y)2]
= (3x – 2y) (9x2 + 6xy + Ay2)

18. (iii) a6 – 64 = a6 – 26
= (a2)3 – (22)3 [a2 – b3 = (a- b) + (a2 + ab + b2)]
= (a2 – 22) [(a2)2 + (a2) (22) + (22)2]
= (a + 2) (a – 2) (a4 + 4a2 + 16)
= (a + 2) (a – 2) [(a2)2 + 42 + 8a2 – 4a2]
= (a + 2)(a- 2) [(a2 + 4)2 – (2a)2] {a2 – b2 = (a + b) (a – b)}
= (a + 2) (a – 2) [(a2 + 4 + 2a) (a2 + 4 – 2a)
= (a + 2) (a – 2) (a2 + 2a + 4) (a2 – 2a + 4)

19. Using the formula [a3 + b3 + c3 – 3abc] = (a + b + c) (a2 + b2 + c2 – ab –
bc – ac)
Solution:
(i) x3 + 8y2 + 6xy – 1 = -(-x3 – 8y3 – 6xy + 1)
= – (-x3 – 8y3 + 1 – 6xy)
= -[(-x)3 + (-2y)3 + 1 – 3(x) (2y) (1)]
= -[-x – 2y + 1] [(-x)2 + (-2y)2 + 12 – (-x) (-2y) – (-2y) (1) – (1) (-x)]
= (x + 2y – 1)(x2 + 4y2 + 1 – 2xy + 2y + x)

20. (ii) l3 – 8m3 – 27n3 – 18lmn
= l3 + (-2m)3 + (-3n)2 – 3(l) (-2m) (-3n)
= (l – 2m – 3n) [l2 + (-2m)2 + (-3n)2 -1 (-2m)] – (-2m)(-3n) –
(-3n)(l)
= (l – 2m – 3n) (l2 + 4m2 + 9n2 + 2lm – 6mn + 3ln)