# hard factorisation question help

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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 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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#2

(Original post by

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

**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

Last edited by mqb2766; 1 month ago

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#3

**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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(Original post by

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

**mqb2766**)What happens when you add or subtract the two factors of the "constant" term?

(Original post by

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.

**RDKGames**)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.

thank you that makes more sense

so the final answer I got is (x-y-1)(x+y-2z-3)

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#5

(Original post by

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)

**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)

(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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(Original post by

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)

**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)

(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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#7

(Original post by

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

**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

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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(Original post by

What is the original question?

But the

x^2 - ()

looks like it could be the difference of two squares?

**mqb2766**)What is the original question?

But the

x^2 - ()

looks like it could be the difference of two squares?

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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#9

(Original post by

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)]

**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)]

[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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(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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#11

(Original post by

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

**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

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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