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Svetach [21]
3 years ago
10

Factorise completely

Mathematics
2 answers:
AleksAgata [21]3 years ago
8 0

Answer:

a) (9a-4)(x-3b)

b) (5a-4)(2b-3)

c) (x-3y)(2h-k)

Step-by-step explanation:

9a(x – 3b ) - 4 (x – 3b) = (9a-4)(x-3b)

Since both 9a and -4 are multiplied by (x-3b), they can be put in a bracket too.

5a(2b – 3 ) - 4 (2b – 3) = (5a-4)(2b-3)

Since both 5a and -4 are multiplied by (2b-3), they can be put in a bracket too.

x( 2h-k )+ 3y ( k – 2h ) = x(2h-k) - 3y (-k + 2h)

We can multiply + 3y ( k – 2h ) by (-1) to switch up the arithmetic operations.

x(2h-k) - 3y (-k + 2h) = x(2h-k) -3y(2h-k)

Since both x and -3y are multiplied by (2h-k), they can be put in a bracket too.

x(2h-k) -3y(2h-k) = (x-3y)(2h-k)

ycow [4]3 years ago
5 0

Answer:

a) 9a(x - 3b) - 4(x - 3b) = (9a - 4)(x - 3b)

b) 5a(2b - 3) - 4(2b - 3) = (5a - 4)(2b - 3)

c) x(2h - k) + 3y(k - 2h)

= x(2h - k) - 3y(2h - k)

= (x - 3y)(2h - k)

Step-by-step explanation:

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