According to the Fundamental Theorem of Algebra, which polynomial function has exactly 8 roots? ORX)-(3x? -x-5)(2x-5
1 x) = (3x* +2x)"
o 9x)= (4x2-7)
OM) - Bx - 4y - 1) (3,2-4

Explanation:

For each expression in the parenthesis of choice A, focus only on the leading terms. Those would be the terms to the very left. The leading terms 3x^2 and 2x^6 multiply to 6x^8. This is the leading term of the final expansion and it's of degree 8. Any degree 8 polynomial will have 8 roots according to the fundamental theorem of algebra. Some of those roots may be complex numbers, or they may be purely real numbers.

Choice B can be ruled out because the leading term 3x^4 is raised to the 4th power to get (3x^4)^4 = 81x^16 as the leading term of the final expansion. This polynomial has 16 roots, but we want 8 roots instead.

Choice C is a similar story to choice B. In this case, we have (4x^2)^3 = 64x^6 as the leading term when you expand everything out. This 6th degree polynomial has 6 roots.

And finally, choice D can be ruled out as well. We follow the same steps as choice A. The leading terms from each parenthesis group multiply to 6x^8*3x^2 = 18x^10 showing we have 10 roots (due to the 10th degree polynomial).

the answer to the question is a

step-by-step explanation:

it has the same slope and the same y-intercept.

y + 11 = -7/9(x - 18)

y + 11 =-7/9x + 14

b

y = -7/9x + 3

y-intercept (0, 3)

answer: dead jah

step-by-step explanation:

bullet + jah= dead jah