To factor the cubic polynomial $x^3 - 7x^2 + 11x - 15$, we proceed as follows:

Step 1: Use the Rational Root Theorem

The Rational Root Theorem suggests that any possible rational root of the polynomial must divide the constant term ($-15$) and be a factor of the leading coefficient ($1$). Therefore, the possible rational roots are:

$\pm 1, \pm 3, \pm 5, \pm 15$

Step 2: Test possible roots

We test the possible roots by substituting them into the polynomial.

Testing $x = 3$:

$P(3) = (3)^3 - 7(3)^2 + 11(3) - 15 = 27 - 63 + 33 - 15 = 0$ So, $x = 3$ is a root.

Step 3: Perform synthetic division

We divide $x^3 - 7x^2 + 11x - 15$ by $x - 3$ using synthetic division:

3 & 1 & -7 & 11 & -15 \\ & & 3 & -12 & -3 \\ \hline & 1 & -4 & -1 & 0 \\ \end{array}$$ The quotient is $$x^2 - 4x - 1$$, and the remainder is $$0$$, confirming the factorization: $$x^3 - 7x^2 + 11x - 15 = (x - 3)(x^2 - 4x - 1)$$ ### Step 4: Factor $$x^2 - 4x - 1$$ further We solve $$x^2 - 4x - 1 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 1$$, $$b = -4$$, and $$c = -1$$: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2(1)} = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$$ ### Final Factorization Thus, the fully factored form of $$x^3 - 7x^2 + 11x - 15$$ is: $$(x - 3)(x - (2 + \sqrt{5}))(x - (2 - \sqrt{5}))$$ Alternatively, it can be written as: $$(x - 3)(x^2 - 4x - 1)$$ Would you like more details on the steps or any specific part explained further? --- ### Related Questions: 1. How does the Rational Root Theorem work in finding possible roots? 2. Can synthetic division handle higher-degree polynomials effectively? 3. What is the significance of the discriminant in the quadratic formula? 4. How can we verify if a factorization is correct? 5. How do we factorize a cubic polynomial without rational roots? ### Tip: When factoring cubic polynomials, always start by checking for simple rational roots using substitution—it can save significant effort!