This is easy but i'm lazy someone explain how to find the solution to an equation like (x+1)(2x-3) or (x-8)^2 IM GIVING 65 POINTTSS

make equation = 0

Step-by-step explanation:

In first case your equation is factored so just = 0

(x+1)(2x-3) = 0

Product is zero if any factor is zero so (x+1) = 0, (2x-3) = 0

Solve for x: x= -1, x= 3/2

If  (x-8)^2 is given than you need to factor it first and then do the same as in first example.

(x-8)^2 = (x-8)(x-8)

(x-8)=0 , x=8

This is double zero.

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."