Attached a solution and showed work.
Answer:
![x^2-100=(x-10)(x+10)](https://tex.z-dn.net/?f=x%5E2-100%3D%28x-10%29%28x%2B10%29)
Explanation:
Given the expression:
![x^2-100](https://tex.z-dn.net/?f=x%5E2-100)
This can be a difference of two squares like:
![x^2-10^2](https://tex.z-dn.net/?f=x%5E2-10%5E2)
and this is:
Answer:
10
Step-by-step explanation:
The absolute value always returns a positive value, that is
| - a | = | a | = a , thus
| 5 - 15 | = | - 10 | = 10
To factorize a polynomial you need to find the zeros, and then you can write the polynomial as a product of bynomials.
<h3>
How to factor a polynomial?</h3>
For a polynomial of the form:
![p(x) = A_n*x^n + A_{n-1}*x^{n-1} + ... + A_1*x + A_0](https://tex.z-dn.net/?f=p%28x%29%20%3D%20A_n%2Ax%5En%20%2B%20A_%7Bn-1%7D%2Ax%5E%7Bn-1%7D%20%2B%20...%20%2B%20A_1%2Ax%20%2B%20A_0)
We define the zeros as the values of x such that:
p(x) = 0.
On the above case, the degree of the polynomial is n (the maximum exponent). Then the number of zeros of the polynomial is n (same as the degree).
Once we know the n zeros, we can factorize the polynomial as:
![p(x) = A_n*(x - x_1)*(x - x_2)*...*(x - x_n)](https://tex.z-dn.net/?f=p%28x%29%20%3D%20A_n%2A%28x%20-%20x_1%29%2A%28x%20-%20x_2%29%2A...%2A%28x%20-%20x_n%29)
For example, for the polynomial:
![p(x) = 2x^2 - 8](https://tex.z-dn.net/?f=p%28x%29%20%3D%202x%5E2%20-%208)
The two zeros are x = 2 and x = -2, then it can be rewriten as:
![p(x) = 2*(x + 2)*(x - 2)](https://tex.z-dn.net/?f=p%28x%29%20%3D%202%2A%28x%20%2B%202%29%2A%28x%20-%202%29)
If you want to learn more about polynomials:
brainly.com/question/4142886
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