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