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The given problem is:

When \( ax^3 - x^2 + 2x + b \) is divided by \( x - 1 \), the remainder is 10. When it is divided by \( x - 2 \), the remainder is 51. We are asked to find the values of \( a \) and \( b \).

### Approach:

By the **remainder theorem**, if a polynomial \( f(x) \) is divided by \( x - c \), the remainder is \( f(c) \). Applying this theorem to the problem:

1. **For division by \( x - 1 \):**
   \[
   f(1) = 10
   \]
   Substituting into the polynomial \( f(x) = ax^3 - x^2 + 2x + b \):
   \[
   f(1) = a(1)^3 - (1)^2 + 2(1) + b = a - 1 + 2 + b = a + b + 1
   \]
   So, \( a + b + 1 = 10 \), which simplifies to:
   \[
   a + b = 9 \quad \text{(Equation 1)}
   \]

2. **For division by \( x - 2 \):**
   \[
   f(2) = 51
   \]
   Substituting into the polynomial:
   \[
   f(2) = a(2)^3 - (2)^2 + 2(2) + b = 8a - 4 + 4 + b = 8a + b
   \]
   So, \( 8a + b = 51 \quad \text{(Equation 2)} \).

### Solving the system of equations:

We now have the system:
1. \( a + b = 9 \)
2. \( 8a + b = 51 \)

From the first equation, solve for \( b \):
\[
b = 9 - a
\]

Substitute this into the second equation:
\[
8a + (9 - a) = 51
\]
\[
8a - a + 9 = 51
\]
\[
7a + 9 = 51
\]
\[
7a = 42
\]
\[
a = 6
\]

Substitute \( a = 6 \) back into the first equation:
\[
6 + b = 9
\]
\[
b = 3
\]



When ax^3 - x^2 + 2x + b is divided by x - 1, the remainder is 10. When it is divided by x - 2, the remainder is 51. Find a and b.

Learn how to find the coefficients a and b in the polynomial ax^3 - x^2 + 2x + b using the remainder theorem. The problem involves applying algebraic concepts to determine a system of equations based on the given remainders. Suitable for students in Grades 9-10.

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