The test of rational roots let us check the possible rational roots of a polynomial.
- The possible rational roots are: [tex]\±1, \±3, \±5, \±9, \±15[/tex] and [tex]\±45[/tex].
- The actual roots are -3 and 3
Possible Rational Roots
The polynomial is given as:
[tex]f(x) = x^4 - 4x^3 - 4x^2 + 36x - 45[/tex]
The leading term of the polynomial is: 1.
The factors of 1 is [tex]\±1[/tex]
The constant term of the polynomial is: -45
The factors of -45 are [tex]\±1, \±3, \±5, \±9, \±15, \±45[/tex]
So, the possible rational roots (r) are:
[tex]r = \frac{\±1, \±3, \±5, \±9, \±15, \±45}{\±1}[/tex]
Split
[tex]r = \frac{\±1}{\±1}, \frac{\±3}{\±1}, \frac{\±5}{\±1}, \frac{\±9}{\±1}, \frac{\±15}{\±1}, \frac{\±45}{\±1}[/tex]
Simplify each division
[tex]r = \±1, \±3, \±5, \±9, \±15, \±45[/tex]
The Actual Roots
We have:
[tex]f(x) = x^4 - 4x^3 - 4x^2 + 36x - 45[/tex]
Expand
[tex]f(x) = x^4 -4x^3 + 5x^2 - 9x^2 + 36x - 45[/tex]
Factorize
[tex]f(x) = x^2(x^2 -4x + 5) - 9(x^2 -4x + 5)[/tex]
Factor out [tex](x^2 -4x + 5)[/tex]
[tex]f(x) = (x^2 - 9)(x^2- 4x + 5)[/tex]
Express 9 as [tex]3^2[/tex]
[tex]f(x) = (x^2 - 3^2)(x^2- 4x + 5)[/tex]
Express [tex]x^2 - 3^2[/tex] as difference of two squares
[tex]f(x) = (x - 3)(x+3)(x^2- 4x + 5)[/tex]
[tex](x^2 -4x + 5)[/tex] cannot be factorized.
So, the actual roots are:
[tex]x - 3 = 0\ or\ x + 3 = 0[/tex]
Solve for x
[tex]x = 3\ or\ x = -3[/tex]
Hence, the actual roots are -3 and 3
Read more about rational and actual roots at:
https://brainly.com/question/19553416
