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Fynjy0 [20]
3 years ago
7

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

Mathematics
2 answers:
gregori [183]3 years ago
5 0
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
oee [108]3 years ago
3 0

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

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