When a polynomial is divided by (x+2), the remainder is -19. When the same polynomial is divided by (x-1), the remainder is 2. Determine the remainder when the polynomial is divided by (x-1)(x+2).
If p(x) is the polynomial, then you have:
p(x) = (x+2)q1(x) - 19
for some poynomial q1(x). You see that the remainder of -19 is the value of
p(x) at x = -2. We also have:
p(x) = (x-1)q2(x) + 2
Therefore p(1) = 2.
Then if you divide p(x) by (x-1) (x+2), the remainder will be a first degree polynomial, so we have:
p(x) = (x-1)(x+2)q3(x) + r(x)
Then if you put x = 1 in here and use that p(1) = 2, you find:
r(1) = 2
Putting x = -2 and using that
p(-2) = -19 yields:
r(-2) = -19
These two values of r(x) fix r(x) as
r(x) is of first degree. We have:
r(x) = (-19)/(-3) (x-1) + 2/3 (x+2) =
7 x - 5
To determine the remainder when a polynomial is divided by a quadratic expression like (x-1)(x+2), we can use the Remainder Theorem.
First, let's represent the polynomial as P(x). We know that when P(x) is divided by (x+2), the remainder is -19, and when it is divided by (x-1), the remainder is 2.
When P(x) is divided by (x+2), we get a quotient Q(x) and a remainder R1. So, we have:
P(x) = Q(x)(x+2) + R1
Similarly, when P(x) is divided by (x-1), we get a quotient Q'(x) and a remainder R2. So, we have:
P(x) = Q'(x)(x-1) + R2
We need to determine the remainder R when P(x) is divided by (x-1)(x+2), which means we need to express P(x) in terms of this quadratic expression.
To find R, we can start by simplifying the expressions for R1 and R2:
R1 = P(x) - Q(x)(x+2)
R2 = P(x) - Q'(x)(x-1)
Since we know that R1 = -19 and R2 = 2, we can set up the equations:
-19 = P(x) - Q(x)(x+2)
2 = P(x) - Q'(x)(x-1)
Now, let's solve these equations simultaneously to find the values of P(x), Q(x), and Q'(x).