Question

Factorise
a^2+2ab-ac-3b^2+5bc-2c^2​

Answer 1

Step-by-step explanation:

The expression is

a2+2ab−ac−3b2+5bc−2c2a2+2ab−ac−3b2+5bc−2c2

Let a = (b-c)

(b−c)2+2b(b−c)−(b−c)c−3b2+5bc−2c2(b−c)2+2b(b−c)−(b−c)c−3b2+5bc−2c2

Or, b2−2bc+c2+2b2−2bc−bc+c2−3b2+5bc−2c2b2−2bc+c2+2b2−2bc−bc+c2−3b2+5bc−2c2

Or, 3b2−5bc+2c2−3b2+5bc−2c23b2−5bc+2c2−3b2+5bc−2c2

Or, 0

(How I know this?

- Seriously hit and trial for 15 minutes using different values.)

Then,

(a-b+c) must be one of the factors

a2+2ab−ac−3b2+5bc−2c2a2+2ab−ac−3b2+5bc−2c2

We have to leave square terms as it is and try to take in product form with (a-b+c)

Or, a2−ab+ac−3b2+3ab+3bc−2c2−2ac+2bca2−ab+ac−3b2+3ab+3bc−2c2−2ac+2bc

Or, a(a−b+c)+3b(−b+a+c)−2c(c+a−b)

Mark me as brilliant

Answer 2

Step-by-step explanation:

is there is no value of a,b,c ....?