8 The function f is defined as \( \mathrm{f}(x)=3^{2 x}-28\left(3^{x}\right)+27, x \in \mathbb{R} \). a Write \( \mathrm{f}(x) \) in the form \( \left(3^{x}-a\right)\left(3^{x}-b\right) \), where \( a \) and \( b \) are real constants. b Hence find the two roots of \( \mathrm{f}(x) \).

To solve the problem, we start with the function given: \[ f(x) = 3^{2x} - 28(3^x) + 27 \] ### Part (a) We can make a substitution to simplify the expression. Let \( y = 3^x \). Then, we have: \[ f(x) = y^2 - 28y + 27 \] Next, we need to factor the quadratic expression \( y^2 - 28y + 27 \). We look for two numbers that multiply to \( 27 \) (the constant term) and add up to \( -28 \) (the coefficient of \( y \)). The numbers that satisfy these conditions are \( -27 \) and \( -1 \): \[ y^2 - 28y + 27 = (y - 27)(y - 1) \] Substituting back \( y = 3^x \), we have: \[ f(x) = (3^x - 27)(3^x - 1) \] Thus, we can write: \[ f(x) = (3^x - 27)(3^x - 1) \] where \( a = 27 \) and \( b = 1 \). ### Part (b) To find the roots of \( f(x) \), we set \( f(x) = 0 \): \[ (3^x - 27)(3^x - 1) = 0 \] This gives us two equations to solve: 1. \( 3^x - 27 = 0 \) 2. \( 3^x - 1 = 0 \) **For the first equation:** \[ 3^x = 27 \] Since \( 27 = 3^3 \), we have: \[ 3^x = 3^3 \implies x = 3 \] **For the second equation:** \[ 3^x = 1 \] Since \( 1 = 3^0 \), we have: \[ 3^x = 3^0 \implies x = 0 \] ### Conclusion The two roots of \( f(x) \) are: \[ \boxed{3} \quad \text{and} \quad \boxed{0} \]