Question

Pre cal/cal master needed

F(x)= 2x^2 + 3x-2 and g(x)= x-2

find equation for f(g(x))

My answer : 2x^2 -8 +3x-8

True answer: 2x^2 -5x

How do we get true answer? I plugged in g(x) into the x's of F(x) but got a waaaaaay different answer.

Answer 1

Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2

Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8

Now that 2(x-2)^2 is done lets move on to 3(x-2).

Use the distributive property and distribute the 3
3(x-2) = 3x - 6

All that is left is the -2

Now lets put it all together
2(x-2)^2 + 3(x-2)-2

2x^2 - 8x + 8 + 3x - 6 - 2

Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0

So all we have left is
2x^2 - 5x