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Sagot :
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]
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]
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