Step 1: Understanding the Question:
We are given that $x+1$ is a factor of a cubic polynomial containing two unknown coefficients, $a$ and $b$. We are also provided with a secondary linear equation involving $a$ and $b$. We must solve for the exact values of both variables.
Step 2: Key Formula or Approach:
Apply the Factor Theorem: if $(x+1)$ is a factor of $f(x)$, then $f(-1) = 0$. This will generate a linear equation. Then, solve the resulting system of two simultaneous linear equations.
Step 3: Detailed Explanation:
Let the polynomial be $f(x) = 2x^3 + ax^2 + 2bx + 1$.
Since $(x + 1)$ is a factor of $f(x)$, according to the Factor Theorem, substituting $x = -1$ must yield 0.
\[ f(-1) = 2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 \]
Simplify the expression:
\[ 2(-1) + a(1) - 2b + 1 = 0 \]
\[ -2 + a - 2b + 1 = 0 \]
\[ a - 2b - 1 = 0 \]
This gives us our first linear equation:
\[ a - 2b = 1 \] --- (Equation 1)
The problem provides a second linear equation:
\[ 2a - 3b = 4 \] --- (Equation 2)
We now solve this system of linear equations. From Equation 1, we can express $a$ in terms of $b$:
\[ a = 2b + 1 \]
Substitute this expression for $a$ into Equation 2:
\[ 2(2b + 1) - 3b = 4 \]
Expand and solve for $b$:
\[ 4b + 2 - 3b = 4 \]
\[ b + 2 = 4 \]
\[ b = 2 \]
Now substitute the value of $b$ back into the expression for $a$:
\[ a = 2(2) + 1 = 4 + 1 = 5 \]
The values of $a$ and $b$ are 5 and 2 respectively.
Step 4: Final Answer:
The values of $a$ and $b$ are 5 and 2.