11) \( 80 x^{3}-16 x^{2}-50 x+10 \)

To factor the polynomial \( 80x^3 - 16x^2 - 50x + 10 \), we can start by looking for common factors and then applying polynomial factorization techniques. 1. **Factor out the greatest common factor (GCF)**: The GCF of the coefficients \( 80, -16, -50, \) and \( 10 \) is \( 2 \). So we can factor out \( 2 \): \[ 2(40x^3 - 8x^2 - 25x + 5) \] 2. **Factor the cubic polynomial**: Now we need to factor \( 40x^3 - 8x^2 - 25x + 5 \). We can use the Rational Root Theorem to test for possible rational roots. The possible rational roots are the factors of the constant term (5) divided by the factors of the leading coefficient (40). Possible rational roots: \( \pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}, \pm \frac{1}{5}, \pm \frac{5}{5}, \pm \frac{1}{10}, \pm \frac{5}{10}, \pm \frac{1}{20}, \pm \frac{5}{20}, \pm \frac{1}{40}, \pm \frac{5}{40} \). Testing \( x = 1 \): \[ 40(1)^3 - 8(1)^2 - 25(1) + 5 = 40 - 8 - 25 + 5 = 12 \quad (\text{not a root}) \] Testing \( x = -1 \): \[ 40(-1)^3 - 8(-1)^2 - 25(-1) + 5 = -40 - 8 + 25 + 5 = -18 \quad (\text{not a root}) \] Testing \( x = \frac{1}{2} \): \[ 40\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right)^2 - 25\left(\frac{1}{2}\right) + 5 = 40\left(\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) - 12.5 + 5 = 5 - 2 - 12.5 + 5 = -4.5 \quad (\text{not a root}) \] Testing \( x = -\frac{1}{2} \): \[ 40\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 - 25\left(-\frac{1}{2}\right) + 5 = 40\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) + 12.5 + 5 = -5 - 2 + 12.5 + 5 = 10.5 \quad (\text{not a root}) \] After testing several values, we find that \( x = \frac{1}{4} \) is a root: \[ 40\left(\frac{1}{4}\right)^3 - 8\left(\frac{1}{4}\right)^2 - 25\left(\frac{1}{4}\right) + 5 = 40\left(\frac{1}{64}\right) - 8\left(\frac{1}{16}\right) - 25\left(\frac{1}{4}\right) + 5 = \frac{40}{64} - \frac{8}{16} - \frac{25}{4} + 5 = \frac{5}{8} - \frac{1}{2} - \frac{25}{4} + 5 = 0 \] 3. **Synthetic division**: Now we can perform synthetic division of \( 40x^3 - 8x^2 - 25x + 5 \) by \( x - \frac{1}{4} \): \[ \begin{array}{r|rrrr} \frac{1}{4} & 40 & -8 & -25 & 5 \\ & & 10 & -2 & -\frac{25}{4} \\ \hline & 40 & 2 & -27 & 0 \\ \end{array} \] This gives us: \[ 40x^2 + 2x - 27 \] 4. **Factoring the quadratic**: Now we need to factor \( 40x^2 + 2x - 27 \). We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 40 \cdot (-27)}}{2 \cdot 40} \] \[ = \frac{-2 \pm \sqrt{4 + 4320}}{80} = \frac{-2 \pm \sqrt{4324}}{80} = \frac{-2 \pm 66}{80} \] This gives us two roots: \[ x = \frac{64}{80} = \frac{4}{5} \quad \text{and} \quad x = \frac{-68}{80} = -\frac{17}{20} \] 5. **Final factorization**: Thus, we can write the polynomial as: \[ 2(40x^3 - 8x^2 - 25x + 5) = 2\left(x - \frac{1}{4}\right)\left(40x^2 + 2x - 27\right) \] The quadratic can be factored as: \[ 40x^2 + 2x - 27 = 40\left(x - \frac{4}{5}\right)\left(x + \frac{17}{20}\right) \] Therefore, the complete factorization of the original polynomial is: \[ 2\left(x - \frac{1}{4}\right)\left(40\left(x - \frac{4}{5}\right)\left(x + \frac{17}{20}\right)\right) \] This is the factorization of the polynomial \( 80x^3 - 16x^2 - 50x + 10 \).