Answer:
a) -x(4[tex]x^{2}[/tex] -x +1)
b) 4[tex]x^{3}[/tex] (x-1)
c) -12[tex]x^{2}[/tex] - 3x + 3
Step-by-step explanation:
Most of it is really just a matter of substituting expressions in. here's how i did it:
f(x) = x-1
g(x) = 4[tex]x^{2}[/tex]
a) (f-g) x [tex]x[/tex]
= (x - 1- 4[tex]x^{2}[/tex]) x [tex]x[/tex]
= x(- 4[tex]x^{2}[/tex]+x-1)
= -x (4[tex]x^{2}[/tex] - x + 1)
*you can take the negative out of the bracket or not, i did though - just makes it neater. They are both right.
b) (fxg) x [tex]x[/tex]
= (4[tex]x^{2}[/tex](x-1)) x [tex]x[/tex]
= 4[tex]x^{3}[/tex] (x-1)
*just bring the x to the front and times it with the 4[tex]x^{2}[/tex]
c) (f+g) x -3
= (x - 1 + 4[tex]x^{2}[/tex]) x -3
= -3 (4[tex]x^{2}[/tex] + x -1)
= -12[tex]x^{2}[/tex] - 3x + 3
*just bring the -3 to the front and times it with the whole bracket to expand since it said evaluate
Hope that helped : )
