19. Find the product of \( f(x) \) and \( g(x) \) when, \( f(x)=x^{2}-3 x+5 \) and \( g(x)=x^{2}-5 x+6 \)\nGroup- \( D[4 \times 4=16] \)\n\( f\left(\frac{1}{2}\right), f(-1) \)

To find the product of \( f(x) \) and \( g(x) \), we first need to multiply the two polynomials: \[ f(x) = x^{2} - 3x + 5, \quad g(x) = x^{2} - 5x + 6. \] Now let's calculate the product \( f(x) \cdot g(x) \): \[ f(x) \cdot g(x) = (x^{2} - 3x + 5)(x^{2} - 5x + 6). \] Using the distributive property (also known as the FOIL method for binomials), we multiply each term in \( f(x) \) with each term in \( g(x) \): 1. \( x^2 \cdot x^2 = x^4 \) 2. \( x^2 \cdot (-5x) = -5x^3 \) 3. \( x^2 \cdot 6 = 6x^2 \) 4. \( -3x \cdot x^2 = -3x^3 \) 5. \( -3x \cdot (-5x) = 15x^2 \) 6. \( -3x \cdot 6 = -18x \) 7. \( 5 \cdot x^2 = 5x^2 \) 8. \( 5 \cdot (-5x) = -25x \) 9. \( 5 \cdot 6 = 30 \) Now, combine the like terms: - \( x^4 \) - \( -5x^3 - 3x^3 = -8x^3 \) - \( 6x^2 + 15x^2 + 5x^2 = 26x^2 \) - \( -18x - 25x = -43x \) - \( +30 \) Putting it all together, we get: \[ f(x) \cdot g(x) = x^4 - 8x^3 + 26x^2 - 43x + 30. \] Now, let's find \( f\left(\frac{1}{2}\right) \) and \( f(-1) \). Calculating \( f\left(\frac{1}{2}\right) \): \[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^{2} - 3\left(\frac{1}{2}\right) + 5 = \frac{1}{4} - \frac{3}{2} + 5. \] Converting to a common denominator (4): \[ \frac{1}{4} - \frac{6}{4} + \frac{20}{4} = \frac{1 - 6 + 20}{4} = \frac{15}{4}. \] Now calculating \( f(-1) \): \[ f(-1) = (-1)^{2} - 3(-1) + 5 = 1 + 3 + 5 = 9. \] Thus, you have \( f\left(\frac{1}{2}\right) = \frac{15}{4} \) and \( f(-1) = 9 \).