anonymous
  • anonymous
1.)When the expression x^3+mx^2+nx-5 is divided by x+2 and x+3, the remainders are 2 and 13 respectively. Find the values of m and n. 2)When P(x)=x^2+ax-b is divided by x-1, the remainder is 6. When P(x) is divided by x+2, the remainder is 3. Find the values of a and b.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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nincompoop
  • nincompoop
for question number 1, try factoring first. See if that'll give you anything
anonymous
  • anonymous
factoring?
anonymous
  • anonymous
how?

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anonymous
  • anonymous
This is factoring : 2 . 3 + 2 . 5 = 2(3+5)
anonymous
  • anonymous
you don't have to factor it referring to the question we can obtain the following equations (x^3+mx^2+nx-5)/x+2 =2 ---- 1. (x^3+mx^2+nx-5)/x+3 = 13 ----2 solve equations 1 and 2 to obtain the values of m and n
anonymous
  • anonymous
x^3+mx^2+nx-5 / x+2 +2 = x^3+mx^2+nx-5 x^3+mx^2+nx-5/x+3 +13= x^3+mx^2+nx-5 Now can you answer ?:)
anonymous
  • anonymous
x^3+mx^2+nx-5 / x+2 +2 = x^3+mx^2+nx-5/x+3 +13
anonymous
  • anonymous
I don't get it ><
anonymous
  • anonymous
oh my logic is incorrect, nevermind
DebbieG
  • DebbieG
Yeah, can't use factoring as you're looking at remainders. I think you want to work backwards with synthetic division.
nincompoop
  • nincompoop
it says that the remainder… so this is about long division or synthetic division my badness, I just woke up earlier and saw cube roots and I thought i can take it like that shall we do this?
nincompoop
  • nincompoop
omg Debbie beat me to it
DebbieG
  • DebbieG
x^3+mx^2+nx-5 divided by x+2 has remainder 2 --> do the synth div, you'll get a quotient expression involving m & n, set the remainder = 1 x^3+mx^2+nx-5 divided by x+3 has remainder 13 --> do the same thing, with r=13. You'll have 2 equations in 2 unknowns. Solve that, and you have m and n. :)
anonymous
  • anonymous
@DebbieG using working backwards in synthetic?
DebbieG
  • DebbieG
lol sorry @nincompoop ... GMTA, huh? :) I've given a very similar problem but only with one unknown and one division & remainder. it's a great little problem! :)
DebbieG
  • DebbieG
@PAULEEnomial maybe that isnt worded well, I don't mean work backwards in the synth. div... just do it as normal, but your result will be in terms of m and n. Now set the remainders = to the value given in each case, and there's your system to solve.
DebbieG
  • DebbieG
Oops, up above ^^ I meant to say "set the remainder = 2" on the first one, of course... not =1.
anonymous
  • anonymous
|dw:1378429991663:dw|
anonymous
  • anonymous
like this?
DebbieG
  • DebbieG
no... hang on...
anonymous
  • anonymous
ha? How?
DebbieG
  • DebbieG
|dw:1378289138187:dw|
anonymous
  • anonymous
can you do the first one? I don't get it eh
DebbieG
  • DebbieG
|dw:1378289352430:dw|
anonymous
  • anonymous
why didn't n-2m+4 turned to -2n+4m-8???
anonymous
  • anonymous
|dw:1378355065822:dw|
DebbieG
  • DebbieG
Sorry, yes, that was my error.... Good catch! :) (I will blame the fact that I was rushing because needed to get my kids out the door to the bus.... yes, that's my story and I'm sticking to it! ;)

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