polynomial help :)

[dianzhang]

Given that the expression P(x) = x^2 - 5x + 7 leaves the remainder whether divided by x-b or x-c, where b does not equal c, show that b+c = 5. Given 4bc = 21 and b is greater than c find the values of b and c


MOD edit: removed poll

[TrueTears]

Given that the expression P(x) = x^2 - 5x + 7 leaves the remainder whether divided by x-b or x-c, where b does not equal c, show that b+c = 5. Given 4bc = 21 and b is greater than c find the values of b and c

leaves a remainder of...?

[dianzhang]

Given that the expression P(x) = x^2 - 5x + 7 leaves the remainder whether divided by x-b or x-c, where b does not equal c, show that b+c = 5. Given 4bc = 21 and b is greater than c find the values of b and c

leaves a remainder of...?

sorry i forgot to add in a word, it leaves the same remainder.

[TrueTears]

Given that the expression P(x) = x^2 - 5x + 7 leaves the remainder whether divided by x-b or x-c, where b does not equal c, show that b+c = 5. Given 4bc = 21 and b is greater than c find the values of b and c

leaves a remainder of...?

sorry i forgot to add in a word, it leaves the same remainder.

yup thought so, just create simultaneous equations like this:

using the remainder theorem P(b) = P(c) ...[1]

You're given 4bc=21 ...[2]

Solve those 2 simultaneously under the constraint that b>c.

Then just add b and c to show that b+c = 5

[dianzhang]

Given that the expression P(x) = x^2 - 5x + 7 leaves the remainder whether divided by x-b or x-c, where b does not equal c, show that b+c = 5. Given 4bc = 21 and b is greater than c find the values of b and c

leaves a remainder of...?

sorry i forgot to add in a word, it leaves the same remainder.

yup thought so, just create simultaneous equations like this:

using the remainder theorem P(b) = P(c) ...[1]

You're given 4bc=21 ...[2]

Solve those 2 simultaneously under the constraint that b>c.

Then just add b and c to show that b+c = 5

ok so i let P(b) = P(c)
and i let b= 21/4c
i ended up with 441/(16c^2) - 105/4c = c^2 - 5c
what should i try to do from here?

[TrueTears]

441/(16c^2) - 105/4c = c^2 - 5c

is that a 105/(4c) or (105/4)c?

[dianzhang]

441/(16c^2) - 105/4c = c^2 - 5c

is that a 105/(4c) or (105/4)c?

105/(4c)

[TrueTears]

right, working from what you have, i'd do this:

441/(16c^2) - 105/(4c)= c^2 - 5c





then just solve the above quartic

[dianzhang]

right, working from what you have, i'd do this:

441/(16c^2) - 105/(4c)= c^2 - 5c





then just solve the above quartic

thats exactly what i did, but this question is meant to be a non calc question.
so any idea what i do from here?

[TrueTears]

i haven't done the arithmetics myself, but i suspect theres a nicer way to combine equations [1] and [2] from earlier, however going on from what you have here:

you'd solve it like any other quartic and guess factors, however using the rational root theorem here helps alot (http://en.wikipedia.org/wiki/Rational_root_theorem)

So factors would be of the form

just happens that if we pick it's a root so is a factor, then long divide etc

[b^3]

You know the two remainders equal each other.




It seems its just a bit of a factorisation trick

From there onwards use and solve simultaneously.






so or
therefore or but
so and

EDIT: Neatened it up with a bit of LaTeX

[TrueTears]

and yeah b^3's post above confirms my suspicions that there was a more elegant way

[dianzhang]

You know the two remainders equal each other.




It seems its just a bit of a factorisation trick

From there onwards use 4bc=21 and solve simultaneously.
4bc=21 (b>c)
c=5-b
4b(5-b)=21
20b-4b²=21
4²-20b+21=0
(2c-3)*(2c-7)=0
so c = 3/2 or c=7/2
therefore b= 7/2 or 3/2 but b>c
so b=7/2 and c =3/2

thanks so much dude, and thank you TT