Question

Find the zeros of each function. State the multiplicity of multiple zeros.

1. y=(x-2)(x+4)

2. y=(x+1)(x-8)(x-9)

3. y=x(x+1)(x+5)

Answer 1

first one
set \[x-2=0\]solve for \(x\) in your head
then set \[x+4=0\] and again solve for \(x\)
others are the same procedure

Answer 2

wait I don't understand.. I'm not even sure what the question is asking

Answer 3

the question is asking which values of \(x\) would you plug in to get zero

Answer 4

in other word, it is asking to solve
\[(x-2)(x+4)=0\] for \(x\)

Answer 5

since it is already factored you have very little work to do
\[x-2=0\iff x=2\] so \(x=2\) is one answer
also \[x+4=0\iff x=-4\] making your two answers \(x=2,x=-4\)

Answer 6

Oh, so that's it? I was thrown off by it asking for zeros.. what does the multiplicity part mean?

Answer 7

In your posted problem, (x-2)(x+4)=0, you'll notice that each of the factor (x-2) and (x+4) shows up just once. We say each has a "multiplicity" of one. Supposing you had
(x-2)(x-2)(x+4)=0 and notice that the factor (x-2) shows up twice. We thus say that (x-2) has a multiplicity of two.

Thinking about this a bit, I realize that the term "multiplicity" applies primarily to the roots (such as x=2, x=-4) more to the factors.

So, if (x-2) shows up twice, the root 2 has a multiplicity of two.

Answer 8

means the exponent of the factor, which is one in every case in your example

Answer 9

\[(x-3)^4\] has a zero of \(3\) with multiplicity \(4\)

Answer 10

ohhh ok. what about y=x^2(x+1)? how would I do that

Answer 11

y = x^2(x + 1)

x^2 = 0 <-- ._. basic math it's zero >.>

x + 1 = 0
-1 -1

x = -1