Question

Q5. Factorise:
(а) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)

Answer 1

Step-by-step explanation:

a) 40 x 14nmp - 441 x 14nmp - 360 x 14nmp

= 14nmp x (40 - 441 - 360)

= -10654nmp

b) 2 x 0

= 0

Answer 2

a) To factorize 14m^5n^4p^2 - 42m^7n^3p^7 - 70m^6n^4p^3, we can take out the common factor of 14m^5n^3p^2 to get:

14m^5n^3p^2(1 - 3m^2np^5 - 5mnp)

b) To factorize 2a^2(b^2 - c^2) + b^2(2c^2 - 2a^2) + 2c^2(a^2 - b^2), we can first simplify each term using the difference of squares identity:

2a^2(b + c)(b - c) + 2b^2(c - a)(c + a) - 2c^2(b + a)(b - a)

Then, we can factor out the common factor of (b + c) to get:

2(b + c)(a^2(b - c) + b^2(c - a) - c^2(b - a))

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