Rational root theorems are used to determine the potential roots of a polynomial function
The function is given as:
[tex]f(x) = 2x^2 + 2x - 24[/tex]
Expand
[tex]f(x) = 2x^2 + 8x - 6x - 24[/tex]
Expand
[tex]f(x) = 2x(x + 4) - 6(x + 4)[/tex]
Factor out x + 4
[tex]f(x) = (2x -6)(x + 4)[/tex]
Set to 0
[tex](2x -6)(x + 4) = 0[/tex]
Split
[tex](2x -6) = 0\ or\ (x + 4) = 0[/tex]
Make x the subject in both cases
[tex]2x = 6\ or\ x= -4[/tex]
Solve for x
[tex]x = 3\ or\ x= -4[/tex]
So, the actual roots are 3 and -4
Recall that, the function is:
[tex]f(x) = 2x^2 + 2x - 24[/tex]
For a function,
[tex]f(x) = px^n + ax^{n-1} +.......+q[/tex]
According to rational root theorem, the roots of f(x) are:
[tex]Roots = \pm \frac{Factors\ of\ q}{Factors\ of\ p}[/tex]
The roots of 24 are:
[tex]24 = \pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24[/tex]
The roots of 2 are:
[tex]2 = \pm1, \pm2[/tex]
So, the possible roots are:
[tex]Roots = \frac{\pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24}{\pm 1, \pm 2}[/tex]
This gives:
[tex]Roots = \pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24,\pm0.5, \pm1, \pm 1.5, \pm 2, \pm 3, \pm 4, \pm6\pm 12[/tex]
Remove repetitions
[tex]Roots = \pm0.5, \pm1, \pm 1.5, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24[/tex]
Read more about rational root theorem at:
https://brainly.com/question/11015301
