Question

2a²+2b²+2c²-4ab-5ac+5bc​ please please i need help to factorize this quadratic expression in its irreducible factors. I have not had much luck​

Answer 1

Answer:

চ(a-b)^2+(b-c)^2+(c-a)^2

=a^2–2ab+b^2+b^2-2bc+c^2+c^2–2ca+a^2

=2a^2+2b^2+2c^2–2ab-2bc-2ca

=2(a^2+b^2+c^2-ab-bc-ca)

From the above relation , If a^2+b^2+c^2-ab-bc-ca=0

Then (a-b)^2+(b-c)^2+(c-a)^2=0

(a-b)^2 is positive for any value of a and b. And it is true for (b-c) and (c-a) also. Addition of three positive value will be 0 only when they individualy is 0

So, (a-b)^2=0 , or, a-b=0, or, a=b

Similarly (b-c)^2=0 or, b=c

So, a=b=c

So, a:b:c=1:1:1

Step-by-step explanation: