Respuesta :
#1 p(x)=(x-3)*(x-5)*(x-1)=x3-9x2+23x-15
You try the next one...a fractional root would look like (x-3/2)
Using the Factor Theorem, it is found that an example of a polynomial of degree at least 3 where one or more of the roots are fractions is given by:
- [tex]f(x) = x^3 - \frac{7}{2}x^2 + \frac{1}{2}x - 1[/tex]
Factor Theorem:
- The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:
[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]
- In which a is the leading coefficient.
In this problem:
- A polynomial of degree 3 will be built, hence it will have 3 roots.
- A leading coefficient of a = 1 is supposed.
- Two non-fractional roots, which are x = 1 and x = 2, hence [tex]x_1 = 1, x_2 = 2[/tex].
- The fractional root is [tex]x_3 = \frac{1}{2}[/tex].
Hence:
[tex]f(x) = (x - x_1)(x - x_2)(x - x_3)[/tex]
[tex]f(x) = (x - 1)(x - 2)\left(x - \frac{1}{2}\right)[/tex]
[tex]f(x) = (x^2 - 3x + 2)\left(x - \frac{1}{2}\right)[/tex]
[tex]f(x) = x^3 - \frac{7}{2}x^2 + \frac{1}{2}x - 1[/tex]
To learn more about the Factor Theorem, you can take a look at https://brainly.com/question/24380382
