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1. Given That G(-5) = G(2) = 0 And That

G(x) = (x - B)(x + C)(x^2 + 7x - 1)

Where B And C Are Positive Integers, Find B And C.

Help? Where Do I Start?
2. i'll have a bash at it
3. (Original post by nimzy)
GO ON THEN, HAVE A GO.
May I suggest you look up the factor theorem in your textbook? And if I'm not mistaken, that's a Pure 1 question, not pure 2...

Here's a clue:
If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
Replace n with your two values and voila.
4. (Original post by meepmeep)
Last time someone said that around me, I got punched in the face.

Anyway, may I suggest you look up the factor theorem in your textbook? And if I'm not mistaken, that's a Pure 1 question, not pure 2...

Here's a clue:
If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
Replace n with your two values and voila.
its in my pure 2 textbook - and it isnt the factor theorum.
5. (Original post by nimzy)
its in my pure 2 textbook - and it isnt the factor theorum.
Well, if you take the second part of what I said:
If the function equals zero when x takes a certain value n, then (x-n) is a factor of the function.
we have G(-5)=0 and G(2)=0
Comparing with the above, we get (x+5) and (x-2) as factors.
Comparing to G(x) = (x - B)(x + C)(x^2 + 7x - 1), we find that (x+5) must be referring to C and (x-2) must be referring to (x-B). Therefore, C=5 and B=2.
6. (Original post by meepmeep)
Well, if you take the second part of what I said:

we have G(-5)=0 and G(2)=0
Comparing with the above, we get (x+5) and (x-2) as factors.
Comparing to G(x) = (x - B)(x + C)(x^2 + 7x - 1), we find that (x+5) must be referring to C and (x-2) must be referring to (x-B). Therefore, C=5 and B=2.
I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
7. (Original post by nimzy)
I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
Sorry, no can do, as to explain it I would need paper to show how it works. Best go e-mail your maths teacher if he hasn't taught you it. The maths department at your school's good anyway!
8. (Original post by joicey)
its bloody christmas holls why the hell you doing that?????
i need to catch up on maths, a little behind, and i also have a physics exam on jan 12th
9. (Original post by nimzy)
Given That G(-5) = G(2) = 0 And That

G(x) = (x - B)(x + C)(x^2 + 7x - 1)

Where B And C Are Positive Integers, Find B And C.

Help? Where Do I Start?
This is definitely in P1/P2 sumwhere.... if G (a) = 0, then
(x - a) is a factor of G (x).

knowing that the third term (x^2 + 7x - 1) wont factorize to any integer (b^2 - 4ac =/= 0), u know that -5 and 2 must be one of -B and +C. since u know that B and C are both positive integers, it must be in the form x - B, and x - C, so u choose sub in -5 and 2 for B and C such that both end up positive.

well clearly B = 2 and C = 5.
10. answer is B= -5, C= -2 right?
11. (Original post by nimzy)
I HAVE NEVER DONE THAT BEFORE, WE HAVENT COVERED THAT.

PLEASE COULD YOU EXPLAIN, COZ I DONT GET ANY OF THAT AT ALL.
I suggest you ask your teacher then, or at least do a little work yourself.

Also, people are more inclined to help when you do not type in capitals - it comes across as shouting and is arrogant.
12. (Original post by Absolution)
answer is B= -5, C= -2 right?
Erm, B and C are both positive, so you have to take C=5 and B=2 for it to work.
13. (Original post by meepmeep)
Erm, B and C are both positive, so you have to take C=5 and B=2 for it to work.
oh right, ah well i'm rubbish at maths anyway at least i got the numbers
14. this isnt that hard........ u just need to know why B and C are one each of 2 and 5, and which one is which..........

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