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

for question number 1, try factoring first. See if that'll give you anything

Answer 2

factoring?

Answer 3

how?

Answer 4

This is factoring :
2 . 3 + 2 . 5 = 2(3+5)

Answer 5

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

Answer 6

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 ?:)

Answer 7

x^3+mx^2+nx-5 / x+2 +2 = x^3+mx^2+nx-5/x+3 +13

Answer 8

I don't get it ><

Answer 9

oh my logic is incorrect, nevermind

Answer 10

Yeah, can't use factoring as you're looking at remainders. I think you want to work backwards with synthetic division.

Answer 11

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?

Answer 12

omg Debbie beat me to it

Answer 13

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. :)

Answer 14

@DebbieG using working backwards in synthetic?

Answer 15

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! :)

Answer 16

@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.

Answer 17

Oops, up above ^^ I meant to say "set the remainder = 2" on the first one, of course... not =1.