Question

use synthetic division to show that $x$ is a solution of the third - degree polynomial equation, and use the result to factor the polynomial completely. list all real solutions of the equation.
$x^{3}+6x^{2}+11x + 6 = 0;x=-3$
factored:
solutions:

Explanation:

Step1: Set up synthetic division

Write -3 (the given root) to the left. List the coefficients 1, 6, 11, 6 of the polynomial $x^{3}+6x^{2}+11x + 6$.

Step2: Perform synthetic - division

Bring down the first coefficient 1. Multiply -3 by 1 to get -3, add to 6 to get 3. Multiply -3 by 3 to get -9, add to 11 to get 2. Multiply -3 by 2 to get -6, add to 6 to get 0. The result is $x^{2}+3x + 2$.

Step3: Factor the quotient

Factor $x^{2}+3x + 2=(x + 1)(x+2)$.

Step4: Write the factored form of the original polynomial

The factored form of $x^{3}+6x^{2}+11x + 6$ is $(x + 3)(x + 1)(x + 2)$.

Step5: Find the solutions

Set each factor equal to zero: $x+3=0$ gives $x=-3$, $x + 1=0$ gives $x=-1$, $x+2=0$ gives $x=-2$.

Answer:

Factored: $(x + 3)(x + 1)(x + 2)$
Solutions: $x=-3,x=-1,x=-2$