What is an equation of the line that passes through the points (-6, -2) and
(6,8)?

y=5/6x+b

Step-by-step explanation:

use formula (y2-y1)/(x2-x1) to look for the slope:

(8-(-2))/(6-(-6))

=10/12

=5/6 (simplified)

then, use the formula y=mx+b to get the equation by substituting m and x, y to either (-6, -2) or (6, 8):

8=5/6(6)+b

8=5+b

3=b (subtract 5 on both sides)

equation:

y=5/6x+3

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."