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

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

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(+/-)√(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`