Question

Divide: 78a 6b 7 by 5a 6b 4
​

Answer 1

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78ab²

Answer 2

Answer:

Example 12 : Factorise the following.

(i) 21x2y3 + 27x3y2 (ii) a3 – 4a2 + 12 – 3a

(iii) 4x2 – 20x + 25 (iv)

2

– 9

9

y

(v) x4 – 256

Solution : (i) 21x2y3 + 27x3y2

= 3 × 7 × x × x × y × y × y + 3 × 3 × 3 × x × x × x × y × y

= 3 × x × x × y × y (7y + 9x) (Using ab + ac = a (b + c))

= 3x2y2 (7y + 9x)

(ii) a3 – 4a2 + 12 – 3a

= a2 (a – 4) – 3a + 12

= a2 (a – 4) – 3 (a – 4)

= (a – 4) (a2 – 3)

(iii) 4x2 – 20x + 25

= (2x)

2 – 2 × 2x × 5 + (5)2

= (2x – 5)2 (Since a2 – 2ab + b2 = (a – b)

2 )

= (2x – 5) (2x – 5)

(iv)

2

– 9

9

y

=

2

2 – (3) 3

  y

   

If there are two numbers you don’t know, that’s not a problem.

You can use two different variables, one for each unknown number.

An equation involving variables can

be true for all values of the variable

– for example, y + y = 2y (this kind

of equation is usually called an

identity).

Or it can be true for only particular

values of the variable – for example,

2y + 3 = 11, which is true only if y =

4.

Finding the values that make an

equation true is called solving the

eq

Step-by-step explanation:

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