28. \( p(x)=36 x^{4}+36 x^{3}-153 x^{2}-81 x+162 \)

To factor the polynomial \( p(x) = 36x^{4} + 36x^{3} - 153x^{2} - 81x + 162 \), let's follow a systematic approach: ### Step 1: Factor Out the Greatest Common Factor (GCF) First, identify the GCF of all the coefficients: - Coefficients: 36, 36, -153, -81, 162 - GCF is **9** So, factor out 9: \[ p(x) = 9 \left(4x^4 + 4x^3 - 17x^2 - 9x + 18\right) \] ### Step 2: Factor the Quartic Polynomial Next, factor the quartic polynomial \( 4x^4 + 4x^3 - 17x^2 - 9x + 18 \). We look to express it as a product of two quadratic polynomials: \[ (4x^2 + ax + b)(x^2 + cx + d) = 4x^4 + 4x^3 - 17x^2 - 9x + 18 \] Expanding the left side and equating coefficients gives us a system of equations: \[ \begin{cases} 4c + a = 4 & \text{(Coefficient of } x^3) \\ 4d + ac + b = -17 & \text{(Coefficient of } x^2) \\ ad + bc = -9 & \text{(Coefficient of } x) \\ bd = 18 & \text{(Constant term)} \end{cases} \] After testing possible integer values for \( b \) and \( d \) (since \( bd = 18 \)), we find that: - \( b = -9 \) and \( d = -2 \) - Solving the system gives \( a = 0 \) and \( c = 1 \) So, the quartic factors as: \[ (4x^2 - 9)(x^2 + x - 2) \] ### Step 3: Factor Further Now, factor each quadratic: \[ 4x^2 - 9 = (2x - 3)(2x + 3) \quad \text{(Difference of squares)} \] \[ x^2 + x - 2 = (x + 2)(x - 1) \quad \text{(Factored by inspection)} \] ### Final Factored Form Combine all the factors: \[ p(x) = 9(2x - 3)(2x + 3)(x + 2)(x - 1) \] **Therefore, the fully factored form of the polynomial is:** \[ \boxed{p(x) = 9(2x - 3)(2x + 3)(x + 2)(x - 1)} \]