To solve this problem, we need to find the values of \(a\) and \(b\) that satisfy the given conditions and then determine the remainder when \(f(x)\) is divided by \((x - 2)\).
1. Given polynomial:
[tex]\[
f(x) = x^4 - 2x^3 + 3x^2 - ax + b
\][/tex]
2. Conditions provided:
- When \(f(x)\) is divided by \((x - 1)\), the remainder is 5.
- When \(f(x)\) is divided by \((x + 1)\), the remainder is 19.
From the Remainder Theorem, we know that:
[tex]\[
f(1) = 5 \quad \text{and} \quad f(-1) = 19
\][/tex]
3. Substitute \(x = 1\) into \(f(x)\):
[tex]\[
f(1) = 1^4 - 2(1^3) + 3(1^2) - a(1) + b = 1 - 2 + 3 - a + b = 2 - a + b
\][/tex]
We know \(f(1) = 5\), so:
[tex]\[
2 - a + b = 5 \quad \Rightarrow \quad -a + b = 3 \quad \text{(Equation 1)}
\][/tex]
4. Substitute \(x = -1\) into \(f(x)\):
[tex]\[
f(-1) = (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + b = 1 + 2 + 3 + a + b = 6 + a + b
\][/tex]
We know \(f(-1) = 19\), so:
[tex]\[
6 + a + b = 19 \quad \Rightarrow \quad a + b = 13 \quad \text{(Equation 2)}
\][/tex]
5. Solve the system of equations:
[tex]\[
\begin{cases}
-a + b = 3 \quad \text{(Equation 1)} \\
a + b = 13 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
Add the two equations to eliminate \(a\):
[tex]\[
(-a + b) + (a + b) = 3 + 13
\][/tex]
[tex]\[
2b = 16 \quad \Rightarrow \quad b = 8
\][/tex]
Substitute \(b = 8\) back into Equation 2:
[tex]\[
a + 8 = 13 \quad \Rightarrow \quad a = 5
\][/tex]
So, the values are:
[tex]\[
a = 5, \quad b = 8
\][/tex]
6. Find the remainder when \(f(x)\) is divided by \((x - 2)\):
Substitute \(a = 5\) and \(b = 8\) into \(f(x)\):
[tex]\[
f(x) = x^4 - 2x^3 + 3x^2 - 5x + 8
\][/tex]
To find the remainder when \(f(x)\) is divided by \((x - 2)\), use the Remainder Theorem:
[tex]\[
\text{Remainder} = f(2)
\][/tex]
Calculate \(f(2)\):
[tex]\[
f(2) = 2^4 - 2(2^3) + 3(2^2) - 5(2) + 8
\][/tex]
[tex]\[
= 16 - 16 + 12 - 10 + 8
\][/tex]
[tex]\[
= 16 - 16 + 12 - 10 + 8
\][/tex]
[tex]\[
= 10
\][/tex]
Therefore, the remainder when \(f(x)\) is divided by \((x - 2)\) is:
[tex]\[
\boxed{10}
\][/tex]
