Question

What are the zeros of the function? What are their multiplicities? F(x) = 3x^3 + 15x^2 +18x

Answer 1

GCF is 3x

3x(x^2 + 5x +6)=0
3x(x +3)(x +2)=0

3x = 0,  x + 3= 0 ,  x + 2 = 0
   x = 0,    x = -3,   x = -2

Answer 2

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

$x=0, multiplicity\ 1\\x=-2, multiplicity\ 1\\x=-3, multiplicity\ 1$