hard factorisation question help

factorise x^2 - y^2+2yz -2zx -4x +2y +2z + 3

what ive done so far:
= x^2 + (-2z-4)x - [y^2 - (2z+2)y - (2z+3)]
= x^2 + (-2z-4)x - (y+1)[y-(2z+3)]
= ?

I dont know what to do after that
the hint the teacher gave was to take out a factor of x but I dont really understand how

Reply 1

mqb2766

Original post by o_reo

What happens when you add or subtract the two factors of the "constant" term?

(edited 2 years ago)

Reply 2

RDKGames

Study Forum Helper

Original post by o_reo

Not sure about that hint.

But what you have done so far is good because you can treat this as a quadratic in x and look for two numbers (or expressions in this case) that add to give -2z-4 and multiply to give -(y+1)(y-2z-3).

But if you distribute the minus into the first bracket, the term becomes

(-y-1)(y-2z-3)

Look … the two factors do indeed add to give -2z-4 so we found our two expressions.

Reply 3

o_reo

Original post by mqb2766

What happens when you add or subtract the two factors of the "constant" term?

Original post by RDKGames

ohhh I didn't realise you were meant to treat (y+1)(y-(2z+3) as a constant
thank you that makes more sense
so the final answer I got is (x-y-1)(x+y-2z-3)

Reply 4

mqb2766

Original post by o_reo

ohhh I didn't realise you were meant to treat (y+1)(y-(2z+3) as a constant
thank you that makes more sense
so the final answer I got is (x-y-1)(x+y-2z-3)

I was using the term "constant" a bit liberally. But as you're treating the polynomial in x,y,z as a quadratic in x, then that is what you're doing. Obviouosly, the trick is to view the two linear factors in x,y,z like
(x - (y+1))(x + (y-2z-3))
So you're searching for the roots
x = (y+1) and -(y-2z-3)

(edited 2 years ago)

Reply 5

o_reo

Original post by mqb2766

in a similar question I have gotten to the point where
(a-b)[x^2 - (a^2 +2ab +b^2)]
how would I factorise the square brackets further, which expression is acting as the 'constant'?

so far I notice that (a^2 +2ab +b^2) = (a+b)^2

Reply 6

mqb2766

Original post by o_reo

What is the original question?
But the
x^2 - ()
looks like it could be the difference of two squares? So in a sense the method is the same?

(edited 2 years ago)

Reply 7

o_reo

Original post by mqb2766

What is the original question?
But the
x^2 - ()
looks like it could be the difference of two squares?

original question ax^2 -a^3 -a^2b +ab^2 +b^3 -bx^2
what I did:
= (a-b)x^2 - (a^3 -b^3) - (a-b)ab
= (a-b) [x^2 - (a^2+ 2ab +b^2)]

Reply 8

mqb2766

Original post by o_reo

original question ax^2 -a^3 -a^2b +ab^2 +b^3 -bx^2
what I did:
= (a-b)x^2 - (a^3 -b^3) - (a-b)ab
= (a-b) [x^2 - (a^2+ 2ab +b^2)]

So did you take the hint(s) about
[x^2 - (a^2+ 2ab +b^2)]

Its arguably simpler (but similar) than your original question.

(edited 2 years ago)

Reply 9

o_reo

Original post by mqb2766

So did you take the hint(s) about
[x^2 - (a^2+ 2ab +b^2)]

ohh it took me a while to see it I dont know why so the final answer would be
(a-b)(x+a+b)(x-a-b)

thank you again!
I dont know why I always get stuck on the penultimate step as I never see what im suppose to do

Reply 10

mqb2766

Original post by o_reo

You're obviously doing it in the right way, just force yourself to factorize the "constant" term (independent of x) which you spotted was -(a+b)^2.

I guess you originally spotted that a=b was a solution by inspecting the coefficients, so you have a factor of (a-b). Once you had that, you could have written
x^2 = (a+b)^2
to get the other two factors if you wanted to do it a bit more verbose, but the difference of two squares was obviously a goto factorization.