Question

find the zeros of this function. state the multiplicity of multiple zeros

y=(x+2)^2(x-5)^4

Answer 1

What values of x will make factors equal to zero?
x + 2 = 0 or x - 5 = 0
x = -2 or x = 5
Since the factor with -2 zero has a power of 2, the -2 has a multiplicity od 2.
Since the factor with the 5 zero has a power of 4, the 5 has a multiplicity of 4.
The answer then is:
-2, multiplicity 2
5, multiplicity 4

Answer 2

so -2, 2, 5, 4 are the answers

Answer 3

No, only x = -2 and x = 5, but with multiplicities.

Answer 4

Think of it this way, with a simpler problem:
If you simply have
x - 4 = 0
Then the solution is x = 4.

What if you have (x - 4)(x - 4) = 0
Then you write x -4 = 0 or x - 4 = 0
so x = 4 or x = 4.
Since you get the same answer twice, x = 4, you call it x=4, multiplicity 2.

What if you had (x-4)(x-4)(x-4) = 0
Then it's x = 4, multiplicity 3.
If you plug in a 4 in for x, you will get (0)(0)(0) = 0
If you plug in -4, you will not get zero. Instead you'll get
(-4-4)(-4-4)(-4-4) = ?
(-8)(-8)(-8) = 512, so clearly, -4 is not a zero or solution of this equation.

In your case, you have
(x + 2)^2(x - 5)^4 = 0
This is the same as
(x + 2)(x + 2)(x - 5)(x - 5)(x - 5)(x - 5) = 0
Each x + 2 = 0 gives an answer of -2, and you have two of them
Each x - 5 gives an answer of x = 5 and you have 4 of them.
So you have x = -2, multiplicity 2 and x = 5, multiplicity 4
There is no x = 2 or x = -5

Answer 5

o ok i get it a little better

Answer 6

so can you explane this equation y(3x+2)^3(x-5)^5 how you just did

Answer 7

This equation written without parentheses is
y(3x + 2)(3x + 2)(3x + 2)(x - 5)(x - 5)(x - 5)(x - 5)(x - 5) = 0
You have many factors, 9 altogether, but there are only 3 different ones:
y, 3x + 2 and x - 5
When you have an equation that has this form which is several factor multiplied together equal zero, you set each factor equal to zero and solve for the variable.
There is no need to write factors that are equal many times becasue their solution will be the same.
There are 3 different factors here, so set each different factor equal to zero and solve:
y = 0 or 3x + 2 = 0 or x - 5 = 0
Now solve each part
y = 0 or 3x = -2 or x = 5
y = 0 or x = -2/3 or x = 5
There are your three different solutions. Since the factors have different exponents, now you need to write the multiplicity of each solution:
y has power 1, so y = 0 only happens once
(3x + 2) has exponent 3, so y = -2/3 has multiplicity 3
(x - 5) has exponent 5, so x = 5 has multiplicity 5

The final answer is:
y = 0, or y = -2/3 multiplicity 3, or y = 5 muiltiplicity 5