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)?

congruent segments

step-by-step explanation:

“ a line segment that has the endpoints on the circle and passes through the midpoint is called the diameter. the diameter is twice the size of the radius. a line segment that has its endpoints on the circular border but does not pass through the midpoint is called a chord. a circle is the same as 360 small circle up the number 0 . i hope this .

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

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