1. First, let's expand the right side of the equation.
#(x - 4)(x+5)(x + A) + Bx + C#
#color(blue)(x^2 + x - 20)color(red)((x + A)) + Bx + C#
#color(violet)(x^3 + Ax^2 + x^2 + Ax - 20x - 20A) + Bx + C#
2. Rearrange the expression, so that like terms are next to each other (this makes it easier to visualize).
#color(violet)(x^3) color(blue)(+ Ax^2 + x^2) color(green)(+ Ax - 20x + Bx) color(red)( - 20A + C)#
3. Compare the right side of the equation to the left side. Is there any variable that we can find?
#color(violet)(x^3) color(blue)(+ 2x^2) color(green)( - 17x) color(red)( - 17) = color(violet)(x^3) color(blue)(+ Ax^2 + x^2) color(green)(+ Ax - 20x + Bx) color(red)( - 20A + C)#
We can find A!
#color(blue)(2x^2 = Ax^2 + x^2)#
Treat the #x^2# just like any variable. In fact, let's remove it altogether to make solving for #A# easier.
#color(blue)(2 = A + 1)#
Now, it's clear that A = 1. Since we know A, we can plug its value into our equation and solve for another variable.
The right side of the equation is now:
#x^3 + color(blue)(A)x^2 + x^2 + color(blue)(A)x - 20x + Bx -20color(blue)A + C#
#x^3 + color(blue)(1)x^2 + x^2 + color(blue)(1)x - 20x + Bx -20color(blue)(*1) + C#
#x^3 + 2x^2 - 19x + Bx - 20 + C#
Add the left side:
#x^3 + 2x^2 - 17x - 17 = x^3 + 2x^2 - 19x + Bx - 20 + C#
4. We have enough information to solve for both B and C.
#-19 x + Bx = -17x#
#Bx = 2x#
#B = 2#
#-20 + C = -17#
#C = 3#
Hope this helps!
