SOLUTION: Which number is a root of f(x)=x^3+6x^2+9x that has multiplicity 1? a. 3 b. 1 c. 0 d. �3 Do you just substitute one of the numbers for x?

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Question 986182: Which number is a root of f(x)=x^3+6x^2+9x that has multiplicity 1?
a. 3
b. 1
c. 0
d. �3
Do you just substitute one of the numbers for x?
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Replace f(x) with 0, then factor

f(x)=x^3+6x^2+9x
0=x^3+6x^2+9x
x^3+6x^2+9x = 0
x(x^2+6x+9) = 0
x(x+3)(x+3) = 0

Then use the zero product property to break up x(x+3)(x+3) = 0 into these three equations

x=0 or x+3=0 or x+3=0

The first equation x=0 means we have a root of 0. This root is of multiplicity 1

The other two equations are exact duplicates of one another. They both lead to x=-3 being the other root of multiplicity 2.

So in the end, the answer is c. 0

The graph confirms this. Notice how we have a double root at x = -3. It doesn't cross over the x axis as it only touches the x axis here. Contrast that with the root at x = 0 where it crosses through the x axis one time.