Question

factorise: 2a²+a-3ab-b+b²​

Answer 1

Answer:

You can consider your expression as a quadratic polynomial P which unknown is b like this

P(b)=2ab2−(3a−1)b−2a−2

Thence, you just need to factor this polynomial to get what you wanted. You can even change the name of the unknown to "x" if you feel more comfortable with that. You’ll just have to change the name back to "b" at the end.

There are very few values for which the following procedure would not be valid. Actually the only value is when a=0 because in that case, P is not quadratic. So I assume that a≠0 .

So let’s proceed…

You can notice that b=2 is a root. So the second root b′ is such that b+b′=3a−12a ()

So b′=−a+12a

and we have

P(b)=2a(b−2)(b+a+12a)

=(b−2)(2ab+a+1)

which is the solution you gave.

If you didn’t notice that 2 is an obvious root for this polynomial, you could still solve the problem as follows

12aP(b)=b2−(3a−1)2ab−a+1a

=(b−(3a−1)4a)2−((3a−1)216a2+a+1a)

=(b−(3a−1)4a)2−9a2−6a+1+16a2+16a16a2

=(b−(3a−1)4a)2−25a2+10a+116a2

=(b−(3a−1)4a)2−((5a+1)4a)2

=(b−(3a−1)4a−(5a+1)4a)(b−(3a−1)4a+(5a+1)4a)

So 12aP(b)=(b−2)(b+(a+1)2a)

And finally

P(b)=(b−2)(2ab+a+1)