The binomial expansion of \(\sqrt{1 + 4x}\) is given by:

\(1 + 2x - 2x^2 + \cdots\)

Now, consider the expression \((a + bx)(1 + 2x - 2x^2)\).

Expanding this, we have:

\((a + bx)(1 + 2x - 2x^2) = a(1 + 2x - 2x^2) + bx(1 + 2x - 2x^2)\)

\(= a + 2ax - 2ax^2 + bx + 2bx^2 - 2bx^3\)

Collecting terms, the coefficient of \(x\) is \(2a + b\), and the coefficient of \(x^2\) is \(-2a + 2b\).

We are given:

\(2a + b = 3\)

\(-2a + 2b = -6\)

Solving these equations simultaneously:

From \(2a + b = 3\), we have \(b = 3 - 2a\).

Substitute \(b = 3 - 2a\) into \(-2a + 2b = -6\):

\(-2a + 2(3 - 2a) = -6\)

\(-2a + 6 - 4a = -6\)

\(-6a + 6 = -6\)

\(-6a = -12\)

\(a = 2\)

Substitute \(a = 2\) back into \(b = 3 - 2a\):

\(b = 3 - 2(2) = 3 - 4 = -1\)

Thus, \(a = 2\) and \(b = -1\).