Solve kx + 12 = 3 kx for x

This is my question~~ ( r ^ 4 - 2r ^ 3 - 23 r ^ 2 - 4r - 32)÷(r+4) ~~ i must use long division to solve it : / i was wondering if anyone world like to solve it. and if so, share step-by-step explanations! you! ❤

Simplify. −4x^2 (5x^4−3x^2+x−2) −20x^6−12x^4+8x^3−8x^2 −20x^6+12x^4−4x^3+8x^2 −20x^8+12x^4−4x^2+8x −20x^6+12x^4+4x^3−8x^2

Given: abcd is a square. prove: ac ? bd. we are given that abcd is a square. if we consider triangle aeb and triangle aed, we see that side is congruent to side ad because sides of a square are congruent. we know that side ae is congruent to side ae by using the . finally, we know that side de is congruent to side because the diagonals of a square bisect each other. therefore, triangle aeb is congruent to triangle aed by congruency. we see that angle aed and angle aeb are a linear pair, and congruent by cpctc. thus, the measure of these angles will be 90°, and diagonal ac is perpendicular to diagonal bd by the .