I am solving X^4-3x^2+2=0 I factored it to be (x^2-2)(x+1)(x-1) as my book suggested. But why is it three sets of parentheses? How do you know that it's supposed to be three sets of parentheses in opposed to two or four? And how come one of the answers according to my book is plus or minus the square root of two? I see why it might be plus the sq. rt. of two, but I don't get how it could be minus. Thank you for your help clearing this up!
If you factorize the polynomial in two steps, you will see it more clearly.
X^4-3x^2+2
easily factorizes into two factors:
(x²-2)(x²-1)
Each of the above factors factorizes into two others by virtue of the identity
(a²-b²)=(a+b)(a-b)
The factor (x²-2) does not have a rational factor, but it can still be factored as
(x²-2)=(x+√2)(x-√2)
and the second one, as you have already done:
(x²-1)=(x+1)(x-1)
So the polynomial actually has four factors, two of which are irrational, and four corresponds to the degree of the polynomial.
Since
(x+√2)(x-√2)(x+1)(x-1)=0,
it should be quite clear why the four zeroes are where they are.
To understand why there are three sets of parentheses in the factorization of the given polynomial, let's look at each of the factors individually:
1. (x^2-2):
This factor is a quadratic expression and can be further factored using the difference of squares formula: a^2 - b^2 = (a + b)(a - b). In this case, a = x and b = √2. So, (x^2-2) can be factored as (x + √2)(x - √2).
2. (x + 1):
This factor is a linear expression and cannot be factored any further.
3. (x - 1):
Similar to the previous factor, this is also a linear expression and cannot be factored any further.
Now, to address why there are specifically three sets of parentheses:
When we multiply all the factors together, each factor contributes one set of parentheses. Think of it like a multiplication problem: (x^2-2)(x + 1)(x - 1) = (x + √2)(x - √2)(x + 1)(x - 1).
So, there are three sets of parentheses because there are three factors that cannot be factored any further.
Regarding the solution ±√2 mentioned in your book:
By setting each factor equal to zero and solving for x, we can find the values that make the original polynomial equation equal to zero.
Setting each factor equal to zero:
1. x + √2 = 0 ⟹ x = -√2
2. x - √2 = 0 ⟹ x = √2
3. x + 1 = 0 ⟹ x = -1
4. x - 1 = 0 ⟹ x = 1
Thus, we have the solutions: x = -√2, √2, -1, and 1. Note that the ± symbol represents both the positive and negative values.
Specifically, for the factor (x^2-2), when we set it equal to zero and solve, we obtain x = ±√2. So, both plus and minus the square root of two are included in the solution.
I hope this explanation clarifies the factorization of the polynomial and the presence of ±√2 in the solution. Let me know if you have any further questions!