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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anonymous
  • anonymous
is there any solutions for that?
anonymous
  • anonymous
i substitute m to the equation, looks like n=9?
anonymous
  • anonymous
while im using synth div. substituting m=2 then working backwards i came up n=9 XD... so confused here :(

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anonymous
  • anonymous
sorry my bad.. its -9 XD
anonymous
  • anonymous
okay im dyin here XD. it wont work in the first condition only in the 2nd
anonymous
  • anonymous
-7/2 ? did it work in the 2nd condition? to have 13 as a remainder? because i think i cant have 1 m and 2 different n's.
anonymous
  • anonymous
yup.
anonymous
  • anonymous
and that is?
anonymous
  • anonymous
x^3+mx^2+nx-5/x+2 + 2 =x^3+mx^2+nx-5/x and x^3+mx^2+nx-5/x+3+13=x^3+mx^2+nx-5/x Understand ?!
anonymous
  • anonymous
okay just a newbie here and i need solutions or reasons? :)
anonymous
  • anonymous
@nelskie16 :Did you understand : x^3+mx^2+nx-5/x+2 + 2 =x^3+mx^2+nx-5/x and x^3+mx^2+nx-5/x+3+13=x^3+mx^2+nx-5/x ???
anonymous
  • anonymous
@E.ali not even a little...
amistre64
  • amistre64
When the expression x^3+mx^2+nx-5 is divided by x+2 and x+3 the remainders are 2 and 13. Find the values of m and n. Just dig in and start dividing to get 2 equations in m and n to fit .... do you know how to divide polynomials? synthetic division might be easier to run if you can
anonymous
  • anonymous
@amistre64 i can start with that ... hold on
anonymous
  • anonymous
@Hero : OK friend ! But my means was right !:) :D
anonymous
  • anonymous
@amistre64 got 2 equations... on the first one... i got -2n + 4m = 15 2nd one... -3n + 9m = 45 ... then im always stuck with those equations :(
.Sam.
  • .Sam.
Or just do like this, let \[p(x)=x^3+mx^2+nx-5\] when p(-2)=2 and p(-3)=13 You can find both m and n from this p(-2)=2 \[(-2)^3+m(-2)^2+n(-2)-5=2...(1)\] p(-3)=13 \[(-3)^3+m(-3)^2+n(-3)-5=13...(2)\]
anonymous
  • anonymous
remainder theorem?
amistre64
  • amistre64
x^3 + mx^2 + nx -5 0 -2 -2m+4 4m-2n-8 -------------------------- -2 | 1 m-2 -2m+n+4 4m-2n-13 = 2 4m -2n = 15 you did good
amistre64
  • amistre64
now its a basic system of equations run -2n + 4m = 15 -3n + 9m = 45 subsitituion or elimination, or however you like to solve them
anonymous
  • anonymous
system??? okay... wait
anonymous
  • anonymous
do the m and n = 15 / 2 ?
amistre64
  • amistre64
-2n + 4m = 15 -3n + 9m = 45 -2n + 4m = 15 -1n + 3m = 15 -2n + 4m = 15 2n - 6m = -30 -------------- -2m = -15; m = 15/2 -2n + 4(15/2) = 15 -2n + 30 = 15 n = 15/2 as well lets give it a try :) dbl chk them to find out
anonymous
  • anonymous
okay .. wait up
anonymous
  • anonymous
didnt work! i think i did it wrong
amistre64
  • amistre64
it works for me ....
anonymous
  • anonymous
how ? i mean the remainder didnt match up when i substitute 15/2 as m and n.
amistre64
  • amistre64
x^3 +15/2 x^2 +15/2x -5 0 -4/2 -22/2 7 -2 1 11/2 -7/2 2 <-- good x^3 +15/2 x^2 +15/2x -5 0 -6/2 -27/2 18 -3 1 9/2 -12/2 13 <-- good
amistre64
  • amistre64
without knowing how you went about your dbl chk, it hard to see where you might have made an error
anonymous
  • anonymous
how did you get -4/2 ? and in the 2nd one why -6/2?
amistre64
  • amistre64
its easier to add and subtract fractions when you have a like denominator
anonymous
  • anonymous
wait got it!!
amistre64
  • amistre64
-2 = -4/2 etc
anonymous
  • anonymous
sorry my bad.. thank you very much. Sorry for stealing some of your time. BTW thanks for helping.
amistre64
  • amistre64
youre welcome, and good luck ;)
amistre64
  • amistre64
the second one is similar in process to the first; but Sams idea might be a good try

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