Question

According to the Rational Root Theorem, the following are potential roots of f(x) = 2x2 2x â€" 24. â€"4, â€"3, 2, 3, 4 Which are actual roots of f(x)?.

Answer 1

Rational root theorems are used to determine the potential roots of a polynomial function

• The actual roots are 3 and -4
• The potential roots are $\pm0.5, \pm1, \pm 1.5, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24$

The function is given as:

$f(x) = 2x^2 + 2x - 24$

Expand

$f(x) = 2x^2 + 8x - 6x - 24$

Expand

$f(x) = 2x(x + 4) - 6(x + 4)$

Factor out x + 4

$f(x) = (2x -6)(x + 4)$

Set to 0

$(2x -6)(x + 4) = 0$

Split

$(2x -6) = 0\ or\ (x + 4) = 0$

Make x the subject in both cases

$2x = 6\ or\ x= -4$

Solve for x

$x = 3\ or\ x= -4$

So, the actual roots are 3 and -4

Recall that, the function is:

$f(x) = 2x^2 + 2x - 24$

For a function,

$f(x) = px^n + ax^{n-1} +.......+q$

According to rational root theorem, the roots of f(x) are:

$Roots = \pm \frac{Factors\ of\ q}{Factors\ of\ p}$

The roots of 24 are:

$24 = \pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24$

The roots of 2 are:

$2 = \pm1, \pm2$

So, the possible roots are:

$Roots = \frac{\pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24}{\pm 1, \pm 2}$

This gives:

$Roots = \pm1, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24,\pm0.5, \pm1, \pm 1.5, \pm 2, \pm 3, \pm 4, \pm6\pm 12$

Remove repetitions

$Roots = \pm0.5, \pm1, \pm 1.5, \pm2, \pm 3, \pm 4, \pm 6, \pm 8, \pm12,\pm 24$

Read more about rational root theorem at:

brainly.com/question/11015301

Answer 2

3 and -4 dudes and dudettes

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