# hard factorisation question help

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#1
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
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1 month ago
#2
(Original post by o_reo)
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
What happens when you add or subtract the two factors of the "constant" term?
Last edited by mqb2766; 1 month ago
0
1 month ago
#3
(Original post by o_reo)
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

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.
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#4
(Original post by mqb2766)
What happens when you add or subtract the two factors of the "constant" term?
(Original post by RDKGames)

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.
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)
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1 month ago
#5
(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)
Last edited by mqb2766; 1 month ago
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#6
(Original post by mqb2766)
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)
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
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1 month ago
#7
(Original post by o_reo)
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
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?
Last edited by mqb2766; 1 month ago
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#8
(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)]
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1 month ago
#9
(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.
Last edited by mqb2766; 1 month ago
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#10
(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
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1 month ago
#11
(Original post by o_reo)
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
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.
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