Need help! Given polynomial divided by (x - 1),the remainder is 5?

Question

and when this polynmoial is divided ( x - 2) the remainder is 7, and that this polynomial may be written as
( x - 1)(x - 2)Q(x) + Ax + B where Q(x) is a polynomial and A B are numbers, find

i) values of A and B
ii) the remainder when polynomial,not Q(x),is divided by (x - 1)(x - 2)

Answer 1

Let P(x) be your polynomial.
Then P(1) = A + B = 5
P(2) = 2A+B =7
So, solving this system,
A = 2, B = 3.
Finally, P(x)/(x-1)(x-2) = Q(x) + (2x+3)/(x-1)(x-2)
or
P(x) = Q(x)(x-1)(x-2) + 2x+ 3
so the remainder is 2x+3.

Answer 2

we are in a position to precise f(x) as: f(x) = (2x+a million)(x-3)g(x) + h(x), the place h(x) is a primary order polynomial. the the remainder of f(x) while divided with the aid of (2x+a million)(x-3) is h(x). the rest while divided with the aid of (x-3) or (2x+a million) is the the remainder of h(x) while divided with the aid of those words. for this reason all we could desire to do is to discover a linear equation h(x) such that it has a the remainder of 9 while divided with the aid of (x-3) and -5 while divided with the aid of (2x+a million). h(x) = a(x-3) + 9 = ax + (-3a + 9) h(x) = b(2x+a million) -5 = 2bx + (b -5) for some numbers a and b the two equations for h(x) could desire to be the comparable equation so the coefficients of x could desire to be the comparable and the numeric words could desire to be the comparable in the two equations. a=2b -3a + 9 = b - 5, or b+3a = 14 Substituting 2b for a interior the 2d equation provides b + 6b = 14, so b=2 and a=2b=4. h(x)=a(x-3)+9 = 4(x-3) + 9 = 4x - 3 right this is a 2d techniques-set. we initiate out the comparable way, yet we convey h(x) as: h(x)= a(2x+a million) - 5 This ensures that h(x) could have a the remainder of -5 while divided with the aid of (2x+a million) we would prefer to choose for a so as that h(x) could have a the remainder of 9 while divided with the aid of (x-3) .......2a ....----------------------- x-3| 2ax + a -5 .....|2ax - 6a .....------------- ......7a - 5 For the rest to be 9, we could desire to have 7a - 5 = 9 a=2 and h(x) = 2(2x+a million) - 5 = 4x-3

Answer 3

(x-1)*(x-2)
= (x^2 - 3x + 2) + 5

x^2 - 3x + 7

but it doesn't apply to the (x-2) division part. the polynomial should be x^2 - 3x + 9

maybe there is something wrong with your question? or maybe i'm just sleepy...

Answer 4

did i mention that i just sat staring at your question for the first five minutes and then i fell down?