Question

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.

Answer 1

is there any solutions for that?

Answer 2

i substitute m to the equation, looks like n=9?

Answer 3

while im using synth div. substituting m=2 then working backwards i came up n=9 XD... so confused here :(

Answer 4

sorry my bad.. its -9 XD

Answer 5

okay im dyin here XD. it wont work in the first condition only in the 2nd

Answer 6

-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.

Answer 7

yup.

Answer 8

and that is?

Answer 9

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 ?!

Answer 10

okay just a newbie here and i need solutions or reasons? :)

Answer 11

@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
???

Answer 12

@E.ali not even a little...

Answer 13

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

Answer 14

@amistre64 i can start with that ... hold on

Answer 15

@Hero : OK friend !
But my means was right !:)
:D

Answer 16

@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 :(

Answer 17

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
\[(-3)^3+m(-3)^2+n(-3)-5=13...(2)\]

Answer 18

remainder theorem?

Answer 19

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

Answer 20

now its a basic system of equations run

-2n + 4m = 15
-3n + 9m = 45 subsitituion or elimination, or however you like to solve them

Answer 21

system??? okay... wait

Answer 22

do the m and n = 15 / 2 ?

Answer 23

-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

Answer 24

okay .. wait up

Answer 25

didnt work! i think i did it wrong

Answer 26

it works for me ....

Answer 27

how ? i mean the remainder didnt match up when i substitute 15/2 as m and n.

Answer 28

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

Answer 29

without knowing how you went about your dbl chk, it hard to see where you might have made an error

Answer 30

how did you get -4/2 ? and in the 2nd one why -6/2?

Answer 31

its easier to add and subtract fractions when you have a like denominator

Answer 32

wait got it!!

Answer 33

-2 = -4/2 etc

Answer 34

sorry my bad.. thank you very much. Sorry for stealing some of your time. BTW thanks for helping.

Answer 35

youre welcome, and good luck ;)

Answer 36

the second one is similar in process to the first; but Sams idea might be a good try