Weegy: when x = 3/4,
q(3/4) = 6(3/4)^3 + 19(3/4)^2 ? 15(3/4) ? 28 = -26, therefore, 3/4 is not a root for the function q(x) = 6x^3 + 19x^2 ? 15x ? 28.
User: Yuri thinks that es002-1.jpg is a root of the following function.
q(x) = 6x3 + 19x2 – 15x – 28
Explain to Yuri why es002-2.jpg cannot be a root.
Weegy: when x = 3/4,
q(3/4) = 6(3/4)^3 + 19(3/4)^2 ? 15(3/4) ? 28 = -26, therefore, 3/4 is not a root for the function q(x) = 6x^3 + 19x^2 ? 15x ? 28.
User: According to the Rational Root Theorem, the following are potential roots of f(x) = 60x2 – 57x – 18.
mc015-1.jpg, mc015-2.jpg, 3, 6
Which is an actual root of f(x)?
mc015-3.jpg
mc015-4.jpg
3
6
Weegy: According to the Rational Root Theorem, the following are potential roots of f(x) = 60x2 – 57x – 18
-6/5, -1/4, 3, 6; An actual root of f(x) is -1/4.
Solution:
-1/4: f(x) = 60(-1/4)^2 - 57(-1/4) - 18 = 0.
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