Math, asked by amitgupta7024303, 3 months ago

M = a

2 – b2 + 2ab

N = a

2 + 4b2 – 6ab

O = b2 + 6

P =a2 – 4ab

S = -2a2 + b2 - ab + a

Find M + N + O + P + S.​

Answers

Answered by shaikhyaqoobali9
0

Answer:

factors of algebraic expressions. An irreducible factor is a factor that cannot be ... 20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m) ... (p − 5)2 [(a − b)2 = a2 − 2ab + b2].

Answered by rajashreekolawale
0

Answer:

12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

The common factors are 2, 2, 3.

And, 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × 11 × x × y

The common factors are 2, y.

And, 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q

28p2q2 = 2 × 2 × 7 × p × p × q × q

The common factors are 2, 7, p, q.

And, 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x

3x2 = 3 × x × x

4 = 2 × 2

The common factor is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab2 = 2 × 2 × 2 × 3 × a × b × b

12a2b = 2 × 2 × 3 × a × a × b

The common factors are 2, 3, a, b.

And, 2 × 3 × a × b = 6ab

(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x

−4x2 = −1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

The common factors are 2, 2, x.

And, 2 × 2 × x = 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

The common factors are 2, 5.

And, 2 × 5 = 10

(viii) 3x2y3 = 3 × x × x × y × y × y

10x3y2 = 2 × 5 × x × x × x × y × y

6x2y2z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y.

And,

x × x × y × y = x2y2

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