Hint: $\displaystyle \begin{align*} x^2 - 3x + 2 = (x - 1)(x - 2) \end{align*}$...
Let $p(x)$ be a polynomial. When $p(x)$ is divided by $x-1$, it leaves the remainder $4$. When it is divided by $x-2$ it leaves the remainder $5$. Then find the remainder when $p(x)$ is divided by $x^2-3x+2$.
$p(1)=4$
$p(2)=5 $
$\Rightarrow p(x)=x+3.$
[/TEX]
I don't think I am right on guessing p(x)=x+3, If yes how can I divide it by a quadratic polynomial.
Yes I saw that already and the divisors of p(x) coinciding with the factors of the third divisor.
But I don't know what does it mean? Should I multiply, divide, add or subtract 4 and 5?????
$p(x)=(x-1)q_1(x)+4=(x-2)q_2(x)+5=(x-1)(x-2)q_3(x)+ax+b$
What to do with their quotients. I am struggling
Or it's because the problem is so simple it's hard to give a "hint" without telling you exactly how to solve it!
First, since p leaves a constant remainder when divided by a linear term, p must be quadratic: that is,
Further, since dividing p(x) by x- 1 leaves remainder 4, p(1)= 1+ a+ b= 4. Since dividing p(x) by x- 2 leaves remainder 5, p(2)= 4+ 2a+ b= 5.
Adding to my confusion, $(x^4-x^3+4) \text{ leaves the remainder (4) when divided by }
\left(x-1 \right) \text{ and quotient } x^3$.
Here it is quartic expression.
Note that we don't actually have to find $p(x)$, which is good (there are too many possibilities).
Here is what we know:
$p(x) = q_1(x)(x - 1) + 4 = q_2(x)(x - 2) + 5$.
We want to find $r(x)$ such that $p(x) = q(x)(x^2 - 3x + 2) + r(x)$.
Note that $r(x) = ax + b$, since $\text{deg}(x^2 - 3x + 2) = 2$.
Let's just play with this a bit.
$r(x) = p(x) - q(x)(x^2 - 3x + 2)$, so:
$r(x) - 4 = p(x) - 4 - q(x)(x^2 - 3x + 2) = q_1(x)(x - 1) + q(x)(x - 2)(x - 1) = [q_1(x) - q(x)(x - 2)](x - 1)$, so $r(x) - 4$ is divisible by $(x - 1)$. This mean that $r(1) - 4 = 0$, that is:
$a + b = 4$.
Similarly,
$r(x) - 5 = p(x) - 5 - q(x)(x^2 - 3x + 2) = q_2(x)(x - 2) + q(x)(x - 2)(x - 1) = [q_2(x) - q(x)(x - 1)](x - 2)$, so $r(x) - 5$ is divisible by $(x - 2)$. What does this say about what $r(2) - 5$ is?