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
