Which of the number(s) below are potential roots of the function?p(x) = x4 + 22x2 – 16x – 12


Which of the number(s) below are potential roots of the function?q(x) = 6x3 + 19x2 – 15x – 28




–16✔ –202
Which of the numbers below are some potential roots of p(x) = x3 + 6x2 − 7x − 60 according to the rational root theorem?
Using the section of the graph shown, which of these potential roots of the function should you test first?f(x) = 2x3 – 9x2 – 6x + 40
According to the rational root theorem, the numbers below are some of the potential roots of f(x) = 10x3 + 29x2 – 66x + 27. Select all that are actual roots.



0±2±4 ±9±3±6 ±12

Evaluate the function for the given values to determine if the value is a root.
Select all of the following that are potential roots of p(x) = x4 − 9x2 − 4x + 12?

X No, this polynomial has complex roots.No, this polynomial has irrational roots.No, this polynomial has irrational and complex roots.✔ Yes.
–10 –7–531524
Which of the numbers below are some potential roots of p(x) = x3 + 6x2 − 7x − 60 according to the rational root theorem?
Find all real solutions of this equation to answer the question.(6 – 2x)(3 – 2x)x = 40



Yuri thinks that is a root of the following function.q(x) = 6x3 + 19x2 – 15x – 28Explain to Yuri why cannot be a root.

According to the Rational Root Theorem, potential rational roots must be in the form p/q, where p is a factor of the constant term (-28) and q is a factor of the leading coefficient (6). Since 3 is not a factor of 28 and 4 is not a factor of 6, 3/4 cannot be a rational root. Additionally, substituting 3/4 into the function for x does not result in 0, and dividing the polynomial by (x - 3/4) would result in a non-zero remainder, confirming that 3/4 is not a root.
Question text not available
-16
✔ No, this polynomial has complex roots.X No, this polynomial has irrational roots.No, this polynomial has irrational and complex roots.Yes.
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No, this polynomial has complex roots.✔ No, this polynomial has irrational roots.X No, this polynomial has irrational and complex roots.Yes.
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