Given:
#f(x) = 2x-5#
#g(x) = 3x^2-5x+2#
#h(x) = x^3-x#
a) In order to find #g(5x+2)# we must plug in for every value of #x# in #g(x)# the value #(5x+2)#.
#g(x) = 3x^2-5x+2#
#g(5x+2) = 3(5x+2)^2-5(5x+2)+2#
Now we simplify the expression on the right hand side.
#= 3(5x+2)(5x+2)-5(5x+2)+2#
We FOIL #(color(red)(5x)color(blue)(+2))(color(green)(5x)color(purple)(+2))# and distribute #(-5)# to #(5x+2)#
#= 3[(color(red)(5x))(color(green)(5x))+(color(red)(5x))(color(purple)(+2))+(color(blue)(2))(color(green)(5x))+(color(blue)(+2))(color(purple)(+2))]-25x-10+2#
#= 3[25x^2+10x+10x)+4]-25x-8#
#= 75x^2+60x+12-25x-8#
#g(5x+2) = 75x^2+35x+4#
b) For this part, we again substitute the given value of #5# into #h(x)#
#f(x) = 2x-5#
#h(x) = x^3-x#
#f(x) - h(5) = 2x-5 - [(5)^3-(5)]#
#= 2x-5 - 125 + 5#
#f(x) - h(5) = 2x-125#
c) For this part, we need only to perform arithmetic on the given functions:
#f(x) = 2x-5#
#g(x) = 3x^2-5x+2#
#h(x) = x^3-x#
#f(x)[h(x)-g(x)] = (2x-5)[x^3-x-(3x^2-5x+2)]#
Simplifying the expression in the brackets, we get
#= (2x-5)[x^3-x-3x^2+5x-2]#
#= (2x-5)[x^3-3x^2+4x-2]#
Now we need to expand and multiply #(2x-5)# by #[x^3-3x^2+4x-2]#
#=(color(red)(2x)color(blue)(-5))[x^3-3x^2+4x-2]#
We can distribute the terms in #(2x-5)# and multiply each of them by #[x^3-3x^2+4x-2]#
#=(color(red)(2x))[x^3-3x^2+4x-2]+(color(blue)(-5))[x^3-3x^2+4x-2]#
This gives:
=#color(red)(2x^4-6x^3+8x^2-4x)#
#color(blue)(-5x^3+15x^2-20x+10)#
Combining like terms, gives:
#=color(red)(2)x^4 + (color(red)(-6)color(blue)(-5))x^3 + (color(red)(8)color(blue)(+15))x^2 + (color(red)(-4)color(blue)(-20))x+color(blue)(10) #
#f(x)[h(x)-g(x)] = 2x^4-11x^3+23x^2-24x+10#
