22) \( 7 a^{2}-11 a+4 \)

1. Multiply the coefficient of \( a^2 \) (which is \(7\)) by the constant term (which is \(4\)) to obtain: \[ 7 \times 4 = 28. \] 2. Find two numbers that multiply to \(28\) and add to \(-11\) (the coefficient of \(a\)). The two numbers are \(-7\) and \(-4\) because: \[ (-7)(-4) = 28 \quad \text{and} \quad (-7) + (-4) = -11. \] 3. Rewrite the quadratic by splitting the middle term: \[ 7a^2 -11a +4 = 7a^2 -7a -4a +4. \] 4. Factor by grouping: - Group the terms: \[ (7a^2 -7a) + (-4a +4). \] - Factor out the common factor from each group: \[ 7a(a-1) -4(a-1). \] 5. Factor out the common binomial factor \((a-1)\): \[ (a-1)(7a-4). \] Thus, the factorization is: \[ (a-1)(7a-4). \]