Question

question 8
2 pts
solve
$a^3 -9a^2 + 14a = 0$
find all solutions other than 0 and fill in the boxes:
a=0, ,

Explanation:

Step1: Factor out the common term

First, we factor out the common factor \(a\) from the left - hand side of the equation \(a^{3}-9a^{2}+14a = 0\).
Using the distributive property \(ab + ac=a(b + c)\) (in reverse), we get \(a(a^{2}-9a + 14)=0\).

Step2: Factor the quadratic expression

Next, we factor the quadratic expression \(a^{2}-9a + 14\). We need to find two numbers that multiply to \(14\) and add up to \(- 9\). The numbers are \(-2\) and \(-7\) since \((-2)\times(-7)=14\) and \((-2)+(-7)=-9\).
So, \(a^{2}-9a + 14=(a - 2)(a - 7)\).
The equation becomes \(a(a - 2)(a - 7)=0\).

Step3: Find the solutions

According to the zero - product property, if \(abc = 0\), then either \(a = 0\), \(b = 0\) or \(c = 0\).
We already know that \(a = 0\) is a solution. For the other factors:
If \(a-2=0\), then \(a = 2\); if \(a - 7=0\), then \(a=7\).

Answer:

\(a = 0\), \(a=\boldsymbol{2}\), \(a=\boldsymbol{7}\)