Help , help,,,help................plzz..........theorems........got stuck!!!!!?

Question

Q2: When 2x^3 - 4x^2 - 5x - 2 is divided by ( x - 1)(x + 2), the remainder is ax + b. this result may be expressed as the identity.

2x^3 - 4x^2 - 5x - 2 = (x - 1)( x + 2)Q(x) + ax + b.
where Q(x) is the quotient.
(a) State the degree of Q(x).
(b) by substituting suitable values of x, find a and b.

Answer 1

Thank GOD Im done with school......good luck with that one. I have NO clue.....sorry.

Answer 2

2x^3 - 4x^2 - 5x - 2 = (x - 1)( x + 2)Q(x) + ax + b.
where Q(x) is the quotient.
(a) State the degree of Q(x)

Since q(x) is the resut of dividing a 3rd degree polynomial by the 2nd degree polynomial (x-1)(x+2) then Q(x) must be of the 1st degree
.
(b) by substituting suitable values of x, find a and b
If x=1 then 2-4 -5 -2 = -9 = a+b *
If x =-2 then -16 -16 +10 -2 = -24 =-2a+b **
Multiply Eq * by -1 getting 9 = -a-b ***
Add eq ** to eq *** getting -15 =-3a, so a = 5
Substitut a=5 into eq * getting -9=5+b, so b =-14

Answer 3

OK, this is not so bad!
(x-1)*(x+2) is x^2 +x - 2
Divide this into 2x^3 - 4x^2 - 5x - 2
using long division:

the first term in Q(X) is 2x because x^2*2x is 2x^3.
The rest, which I did by hand and you can, too, gives:
Q(X) = 2x-6, and the remainer is 5x-14

The degree of Q(X) which you did not need to do the long division to figure out is 1 (that is, the largest power X is raised to is 1).

(b) a and b are 5 and -14. I'm not sure what it means when it said "by substituting suitable values of x" - I just took the 5 and -14 from the remainder.

Answer 4

The other answers are correct and easy to do. Another approach would be to long divide by (x-1)(x+2)= x^2+x-2 and then look-see what the answers are. It also gives you a way to check the other work.....

Answer 5

a.degree of Q(x) will be 1, two less than f(x)
b.f(x)=2x^3-4x^2-5x-2
f(1)=2-4-5-2=-9=a+b
f(-2)=-16-16+10-2=-24=-2a+b
subtracting -3a=-15
a=5
a+b=-9
5+b=-9
b=-14