Question

According to the Rational Root Theorem, the following are potential roots of f(x) = 60x2 – 57x – 18. Negative six-fifths, Negative one-fourths, 3, 6 Which is an actual root of f(x)? Negative six-fifths Negative one-fourths 3 6

Answer 1

The actual root of the polynomial f(x) = 60x^2 – 57x – 18 is -6/5, as determined by the Rational Root Theorem and testing each potential root.

Step-by-step explanation:

To determine which of the listed values is an actual root of the polynomial f(x) = 60x2 – 57x – 18, we can apply the Rational Root Theorem and then test each potential root.

The Rational Root Theorem suggests that any rational root of the form π/q, where p and q are integers, must have p as a factor of the constant term and q as a factor of the leading coefficient. Given the polynomial, the potential rational roots must be factors of -18 divided by factors of 60.

The potential rational roots would therefore include ±6⅛ (¶6/5), ¶1/4, ±3, and ±6. To test each root, we substitute them into the polynomial and see if the result equals zero. After testing, we find that:

• f(-6/5) = 60(-6/5)2 – 57(-6/5) – 18 = 0, so -6/5 is a root.
• f(-1/4) does not equal zero, so -1/4 is not a root.
• f(3) does not equal zero, so 3 is not a root.
• f(6) does not equal zero, so 6 is not a root.

Therefore, the actual root of the polynomial f(x) = 60x2 – 57x – 18 is Negative six-fifths (-6/5).