Factorise the following sqares trinomials (a) \( 3 x^{2}-7 x-6 \) (b) \( 3 x^{2}-5 x+2 \)

Certainly! Let's factorize each of the given quadratic trinomials step by step. ### (a) \( 3x^{2} - 7x - 6 \) **Step 1: Identify coefficients** - \( a = 3 \) - \( b = -7 \) - \( c = -6 \) **Step 2: Multiply \( a \) and \( c \)** - \( a \times c = 3 \times (-6) = -18 \) **Step 3: Find two numbers that multiply to \(-18\) and add to \(-7\)** - The numbers are \(-9\) and \(2\) because: - \(-9 \times 2 = -18\) - \(-9 + 2 = -7\) **Step 4: Rewrite the middle term using these numbers** \[ 3x^{2} - 9x + 2x - 6 \] **Step 5: Factor by grouping** \[ (3x^{2} - 9x) + (2x - 6) = 3x(x - 3) + 2(x - 3) = (3x + 2)(x - 3) \] **Final Factored Form:** \[ (3x + 2)(x - 3) \] --- ### (b) \( 3x^{2} - 5x + 2 \) **Step 1: Identify coefficients** - \( a = 3 \) - \( b = -5 \) - \( c = 2 \) **Step 2: Multiply \( a \) and \( c \)** - \( a \times c = 3 \times 2 = 6 \) **Step 3: Find two numbers that multiply to \(6\) and add to \(-5\)** - The numbers are \(-2\) and \(-3\) because: - \(-2 \times -3 = 6\) - \(-2 + (-3) = -5\) **Step 4: Rewrite the middle term using these numbers** \[ 3x^{2} - 2x - 3x + 2 \] **Step 5: Factor by grouping** \[ (3x^{2} - 2x) + (-3x + 2) = x(3x - 2) -1(3x - 2) = (3x - 2)(x - 1) \] **Final Factored Form:** \[ (3x - 2)(x - 1) \] --- **Summary of Factorized Forms:** - **(a)** \( 3x^{2} - 7x - 6 = (3x + 2)(x - 3) \) - **(b)** \( 3x^{2} - 5x + 2 = (3x - 2)(x - 1) \)