Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero
P(x)= x^3-x^2+x
clearly you have
x(x^2-x+1)
Only one real root, of multiplicity 1.
Two complex roots, also of multiplicity 1.
To factor the polynomial P(x) = x^3 - x^2 + x completely and find its zeros, we can follow a systematic approach known as factoring by grouping.
Step 1: Group the terms:
P(x) = x^3 - x^2 + x
Grouping the terms, we get:
P(x) = (x^3 - x^2) + x
Step 2: Factor out the greatest common factor (GCF) from each group:
In the first group, x^3 - x^2, we can factor out an x^2:
P(x) = x^2(x - 1) + x
Step 3: Factor out the common factor from the entire polynomial:
Now, we can factor out the common factor, which is x:
P(x) = x(x^2 - 1) + x
Step 4: Simplify and factor further:
We can simplify the expression x^2 - 1 by recognizing that it is a difference of squares:
P(x) = x(x - 1)(x + 1) + x
This is the factored form of the polynomial P(x). Now, let's find the zeros and determine their multiplicities.
To find the zeros, we set P(x) = 0:
x(x - 1)(x + 1) + x = 0
Now, we solve for x:
x(x - 1)(x + 1) + x = 0
(x^2 - x)(x + 1) + x = 0
(x^2 - x)(x + 1) = -x
We have two cases to consider to find the zeros of the polynomial:
Case 1: x^2 - x = 0
x(x - 1) = 0
x = 0 or x = 1
Case 2: x + 1 = -x
2x = -1
x = -1/2
So, the zeros of the polynomial P(x) = x^3 - x^2 + x are x = 0, x = 1, and x = -1/2.
To determine the multiplicity of each zero, we can examine the factored form of the polynomial:
P(x) = x(x - 1)(x + 1) + x
The multiplicity of a zero is determined by the power to which it appears in the factored form. In this case, x = 0 appears with a power of 1, x = 1 appears with a power of 1, and x = -1/2 appears with a power of 1.
Therefore, each zero of the polynomial P(x) has a multiplicity of 1.