1. **Problem statement:** Find the remainder when $f(x) = x^3 - 2x^2 - 5x + 6$ is divided by $x - 1$ and then factorize $f(x)$. 2. **Remainder theorem:** When a polynomial $f(x)$ is divided by $x - a$, the remainder is $f(a)$. 3. **Calculate remainder:** Substitute $x = 1$ into $f(x)$: $$f(1) = 1^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$$ The remainder is 0, so $x - 1$ is a factor of $f(x)$. 4. **Factorize $f(x)$:** Since $x - 1$ is a factor, divide $f(x)$ by $x - 1$ using polynomial division or synthetic division. Using synthetic division: - Coefficients: 1 (for $x^3$), -2 (for $x^2$), -5 (for $x$), 6 (constant) - Divide by root $x=1$: | 1 | -2 | -5 | 6 | |---|----|----|---| | | 1 | -1 | -6| | 1 | -1 | -6 | 0 | The quotient is $x^2 - x - 6$. 5. **Factorize quotient:** $$x^2 - x - 6 = (x - 3)(x + 2)$$ 6. **Final factorization:** $$f(x) = (x - 1)(x - 3)(x + 2)$$ --- **Next problem:** Solve $e^x - 6e^{-x} + 1 = 0$. 1. **Rewrite equation:** Multiply both sides by $e^x$ to clear the negative exponent: $$e^{2x} - 6 + e^x = 0$$ 2. **Substitute:** Let $y = e^x$, then $y > 0$ and the equation becomes: $$y^2 + y - 6 = 0$$ 3. **Solve quadratic:** $$y = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ 4. **Find roots:** - $y = \frac{-1 + 5}{2} = 2$ - $y = \frac{-1 - 5}{2} = -3$ (discard since $y = e^x > 0$) 5. **Back-substitute:** $$e^x = 2 \implies x = \ln 2$$ **Final solution:** $$x = \ln 2$$