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1. In class we've been given a problem to try and solve and explain.

we had to do (x+_)(_x+_) where we filled the gaps with any three integers between -4 and 4. We then wrote out all the different possible combinations of these (there was 6) and then expand and simplify them. We then had to add all the expanded quadratics together to form another quadratic, which we then had to factorise. each time this came to (2x-3)(x+1).

We now have to explain why this happens in mathematical terms,and why it only happens for integers between -4 and 4, and I have no idea why.
Please could someone explain why this happens?

thanks
2. (Original post by Boople)
In class we've been given a problem to try and solve and explain.

we had to do (x+_)(_x+_) where we filled the gaps with any three integers between -4 and 4. We then wrote out all the different possible combinations of these (there was 6) and then expand and simplify them. We then had to add all the expanded quadratics together to form another quadratic, which we then had to factorise. each time this came to (2x-3)(x+1).

We now have to explain why this happens in mathematical terms,and why it only happens for integers between -4 and 4, and I have no idea why.
Please could someone explain why this happens?

thanks
There are loads more than 6
3. I think he/she is saying that you pick three numbers between -4 and 4, then substitute them into the expression. Given this, there are 3! = 6 ways of arranging the three numbers.

(Original post by TenOfThem)
There are loads more than 6
4. (Original post by WishingChaff)
I think he/she is saying that you pick three numbers between -4 and 4, then substitute them into the expression. Given this, there are 3! = 6 ways of arranging the three numbers.
Ahhhhhhh

that would make more sense

But wouldn't work

I just tried 3, -2, 1 and did not get (2x-3)(x+1)
5. Yeah, at first I thought this sounded like a nice problem but seems to be lacking extra requirements/structure. I noticed the same, but it is much easier to simply realise that if you pick three positive numbers (1,2,3) clearly there is no way you are going to get a factor of 2 in the x^2 term.

(Original post by TenOfThem)
Ahhhhhhh

that would make more sense

But wouldn't work

I just tried 3, -2, 1 and did not get (2x-3)(x+1)
6. (Original post by WishingChaff)
Yeah, at first I thought this sounded like a nice problem but seems to be lacking extra requirements/structure. I noticed the same, but it is much easier to simply realise that if you pick three positive numbers (1,2,3) clearly there is no way you are going to get a factor of 2 in the x^2 term.
I did a general a,b,c and you get 2(x+1)(something) every time

You are going to get a multiple of 2x^2 because they all happen twice
7. Sorry, but maybe I am confused with the initial premise. My point was that if you take the three numbers (1,2,3) and arrange them in there 6 possible ways (where each term corresponds to the point in the expression we substitute into i.e. (1,3,2) = (x+1)(3x+2) and so on.), it is clear that summing all these terms is not going to produce a factor of 2 in the x^2 term (it will be larger than 2). Thus, it is clear that the sum of these expressions does not equal the claimed expression. i.e. (2x-3)(x+1).

(Original post by TenOfThem)
I did a general a,b,c and you get 2(x+1)(something) every time

You are going to get a multiple of 2x^2 because they all happen twice
8. (Original post by WishingChaff)
Sorry, but maybe I am confused with the initial premise. My point was that if you take the three numbers (1,2,3) and arrange them in there 6 possible ways (where each term corresponds to the point in the expression we substitute into i.e. (1,3,2) = (x+1)(3x+2) and so on.), it is clear that summing all these terms is not going to produce a factor of 2 in the x^2 term (it will be larger than 2). Thus, it is clear that the sum of these expressions does not equal the claimed expression. i.e. (2x-3)(x+1).
I am agreeing with you - we get a multiple of 2 in each co-efficient
9. Ok. Well, guess we will have to wait for the student to correct the question.

(Original post by TenOfThem)
I am agreeing with you - we get a multiple of 2 in each co-efficient
10. (Original post by WishingChaff)
Ok. Well, guess we will have to wait for the student to correct the question.
I just repeated what my teacher told me, I'm as lost as you are on this one
11. Then we must conclude that your teacher is wrong.

(Original post by Boople)
I just repeated what my teacher told me, I'm as lost as you are on this one
12. (Original post by Boople)
I just repeated what my teacher told me, I'm as lost as you are on this one
I thought you said that you had done it

In the OP you described the process as though you had actually worked through it in class
13. (Original post by TenOfThem)
I thought you said that you had done it

In the OP you described the process as though you had actually worked through it in class
Nope, we were meant to work through it at home.

Anyway, my teacher has since said that each final quadratic you get should have one of its factors as (x+1), which I checked and it is correct.
Now as for explaining why I formed a general a,b and c one that came out as 2((a+b+c)x^2+2(a+b+c+ab+bc+ca)x+ (ab+bc+ca))
As (x+1) is a factor f(-1) must equal 0
And when putting x=-1 into the general a,b and c thing I got it came out as 0.
14. (Original post by Boople)
Nope, we were meant to work through it at home.

Anyway, my teacher has since said that each final quadratic you get should have one of its factors as (x+1), which I checked and it is correct.
Now as for explaining why I formed a general a,b and c one that came out as 2((a+b+c)x^2+2(a+b+c+ab+bc+ca)x+ (ab+bc+ca))
As (x+1) is a factor f(-1) must equal 0
And when putting x=-1 into the general a,b and c thing I got it came out as 0.
It does actually factorise with 2(x+1) coming out nicely

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