1. The problem is to redefine the modulus function $F(x) = |x^2 - 8x + 15|$ without the absolute value. 2. First, factor the quadratic inside the absolute value: $$x^2 - 8x + 15 = (x - 3)(x - 5)$$ 3. The expression inside the modulus changes sign at the roots $x=3$ and $x=5$. 4. For $x < 3$, both $(x-3)$ and $(x-5)$ are negative, so their product is positive. Thus, $x^2 - 8x + 15 > 0$ for $x < 3$. 5. For $3 < x < 5$, $(x-3)$ is positive but $(x-5)$ is negative, so the product is negative. Thus, $x^2 - 8x + 15 < 0$ for $3 < x < 5$. 6. For $x > 5$, both $(x-3)$ and $(x-5)$ are positive, so the product is positive. Thus, $x^2 - 8x + 15 > 0$ for $x > 5$. 7. Therefore, the modulus function can be rewritten as a piecewise function: $$F(x) = \begin{cases} x^2 - 8x + 15, & x \leq 3 \\ -(x^2 - 8x + 15), & 3 < x < 5 \\ x^2 - 8x + 15, & x \geq 5 \end{cases}$$ 8. This removes the absolute value by considering the sign of the quadratic expression in each interval. Final answer: $$F(x) = \begin{cases} x^2 - 8x + 15, & x \leq 3 \\ -x^2 + 8x - 15, & 3 < x < 5 \\ x^2 - 8x + 15, & x \geq 5 \end{cases}$$