Question

1. f(x) = x³ + 125

factored form: _____________
solutions: _____________

Explanation:

Step1: Recall Sum of Cubes Formula

The sum of cubes formula is \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\). Here, \(a = x\) and \(b^3=125\), so \(b = 5\) (since \(5^3 = 125\)).

Step2: Apply the Formula to Factor

Substitute \(a = x\) and \(b = 5\) into the sum of cubes formula. So \(x^3+125=x^3 + 5^3=(x + 5)(x^2-5x + 25)\).

Step3: Find the Solutions

To find the solutions, set \(f(x)=0\), i.e., \((x + 5)(x^2-5x + 25)=0\).

• For the first factor: \(x + 5 = 0\) gives \(x=-5\).
• For the quadratic factor \(x^2-5x + 25 = 0\), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b=-5\), \(c = 25\). The discriminant \(\Delta=b^2-4ac=(-5)^2-4\times1\times25=25 - 100=-75\). So \(x=\frac{5\pm\sqrt{-75}}{2}=\frac{5\pm5i\sqrt{3}}{2}\) (complex solutions). But if we consider real solutions only, the solution is \(x=-5\). If we consider all (real and complex) solutions, they are \(x=-5\), \(x=\frac{5 + 5i\sqrt{3}}{2}\), \(x=\frac{5 - 5i\sqrt{3}}{2}\).

Answer:

Factored Form: \((x + 5)(x^2 - 5x + 25)\)
Solutions: \(x=-5\) (real solution) or \(x=-5\), \(x=\frac{5 + 5i\sqrt{3}}{2}\), \(x=\frac{5 - 5i\sqrt{3}}{2}\) (all solutions)