8. [-/1 Points] DETAILS MY NOTES TGEIALG6 6.3.086. Factor. If an expression is prime, so indicate. (If the expression is not factorable using integers, enter PRIME. \( -14 t^{2}+1+5 t \) SUBMIT ANSWER

We begin with the expression: \[ -14t^2 + 5t + 1 \] It is easier to factor by first factoring out \(-1\): \[ -14t^2 + 5t + 1 = -\left(14t^2 - 5t - 1\right) \] Next, we factor \(14t^2 - 5t - 1\). 1. Multiply \(a \cdot c\) where the quadratic is in the form \(at^2 + bt + c\): \[ a \cdot c = 14 \cdot (-1) = -14 \] 2. We seek two numbers that multiply to \(-14\) and add to \(b = -5\). These numbers are \(-7\) and \(2\) since: \[ -7 \times 2 = -14 \quad \text{and} \quad -7 + 2 = -5 \] 3. Rewrite the middle term using these numbers: \[ 14t^2 - 5t - 1 = 14t^2 - 7t + 2t - 1 \] 4. Factor by grouping: \[ (14t^2 - 7t) + (2t - 1) \] Factor out the greatest common factor from each group: \[ 7t(2t - 1) + 1(2t - 1) \] 5. Factor out the common binomial \((2t - 1)\): \[ (2t - 1)(7t + 1) \] 6. Incorporate the factor of \(-1\) we factored out initially: \[ -\left(2t - 1\right)\left(7t + 1\right) \] Thus, the factorization of the original expression is: \[ - (2t-1)(7t+1) \]