Answer:
\(x=0, multiplicity\ 1\\x=-2, multiplicity\ 1\\x=-3, multiplicity\ 1\)
Step-by-step explanation:
we have
\(f(x)=3x^{3}+15x^{2} +18x\)
To find the zeros of the function equate to zero
\(3x^{3}+15x^{2} +18x=0\)
Factor the term 3x
\(3x(x^{2}+5x+6)=0\)
Solve the quadratic equation
The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to
\(x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}\)
in this problem we have
\(x^{2}+5x+6=0\)
so
\(a=1\\b=5\\c=6\)
substitute in the formula
\(x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(6)}} {2(1)}\)
\(x=\frac{-5(+/-)\sqrt{25-24}} {2}\)
\(x=\frac{-5(+/-)1} {2}\)
\(x=\frac{-5+1} {2}=-2\)
\(x=\frac{-5-1} {2}=-3\)
so
\(x^{2}+5x+6=(x+2)(x+3)\)
substitute
\(3x(x^{2}+5x+6)=3x(x+2)(x+3)\)
The zeros are
\(x=0, multiplicity\ 1\\x=-2, multiplicity\ 1\\x=-3, multiplicity\ 1\)
