Question

HELP!!! i want to solve (fg)(7) but i don't know what to do next! http://bamboodock.wacom.com/doodler/0ff91a79-1a98-4159-baaa-4279279c9658

Answer 1

expand those brackets, simplify, then sub x=7

Answer 2

but i want to multiply right?

Answer 3

so how do i do that?

Answer 4

Or you can substitute first...its a little easier.

Answer 5

how do you do that?

Answer 6

you get the same answer

Answer 7

\[(x^2-3x+2)(2x-4)=x^2(2x-4)-3x(2x-4)+2(2x-4)\]

Answer 8

(49-21+2)(14-4)=(28+2)(10)=(30)(10)=300

Answer 9

ah so i put the 7 in the first place!!!!

Answer 10

yep

Answer 11

if you expand and then substitute, you get the same answer, just takes longer.

Answer 12

ahh was helpful thank you very much! and thank you @UnkleRhaukus for helping me as well!

Answer 13

hmm, yeah, it is quicker to substitute first

Answer 14

but thank so for helping me out as well!!

Answer 15

\[(fg)(x)=(x^2-3x+2)(2x-4)\]\[\qquad=x^2(2x-4)-3x(2x-4)+2(2x-4)\]\[\quad=2x^3-4x^2-6x^2+12x+4x-8\]\[=2x^3-10x^2+16x-8\]
\[(fg)(7)=2\times 7^3-10\times7^2+16\times7-8\]\[\quad\quad=2\times343-10\times49+112-8\]\[\quad~=686-490+112-8\]\[\quad=196+104\]\[=300\]


PROSS's method is much simpler for this question