If \(x + 1\) is a factor of \(2x^3 + ax^2 + 2bx + 1\) and \(2a - 3b = 4\), find \(a\) and \(b\).
\(a = 5\), \(b = 2\).
Step 1: Apply the Factor Theorem
If \(x + 1\) is a factor, then putting \(x = -1\) in the polynomial should give 0.
Step 2: Substitute \(x = -1\)
Polynomial: \(2x^3 + ax^2 + 2bx + 1\)
\(= 2(-1)^3 + a(-1)^2 + 2b(-1) + 1\)
Step 3: Simplify each term
\((-1)^3 = -1\), so \(2(-1)^3 = -2\).
\((-1)^2 = 1\), so \(a(-1)^2 = a\).
\(2b(-1) = -2b\).
And \(+1\) stays the same.
Step 4: Combine all terms
\(-2 + a - 2b + 1 = 0\)
\(a - 2b - 1 = 0\)
So, \(a = 2b + 1\). (Equation 1)
Step 5: Use the second condition
We are also given: \(2a - 3b = 4\).
Step 6: Substitute \(a = 2b + 1\)
\(2(2b + 1) - 3b = 4\)
\(4b + 2 - 3b = 4\)
\(b + 2 = 4\)
\(b = 2\)
Step 7: Find \(a\)
From Equation (1): \(a = 2b + 1\).
\(a = 2(2) + 1 = 4 + 1 = 5\).
Final Answer: \(a = 5, b = 2\).