I'm having a hard time understandind these, would be appreciated

f(3) = 3^3 - 14*9 + 61*9 - 84 = 0 so 3 is one zero long division of the expression by (x - 3) gives the quadratic x^2 -11x + 28 which = (x - 7)(x-4) so the other zeros are 4 and 7, 3,4 and 7

Basically your answer is 7, 4, and 3

Step-by-step explanation:

5x^3-6x²-59x+12=5x^3-x²-5x²+x-60x+12 =x²(5x-1)-x(5x-1)-12(5x-1) =(5x-1)(x²-x-12) =(5x-1)(x²+3x-4x-12) =(5x-1)(x(x+3)-4(x+3)) =(5x-1)(x+3)(x-4) zeros are

And your answer for this is; 4, -3, 1/5

Hope this helped and good luck :)

(Brainliest is always appreciated)

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)