What conditions would be enough to prove that P is the circumcenter of HJK?

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

what? ?

what type of question is thiss

step-by-step explanation:

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =