The polynomial f(x)=(2x+1)(x+2)(x-3) evaluate f(0) and 2 more questions? 10 points

answer any of these questions you know (mathematicians please answer all 3 pleaseee :) )
a)evaluate f(0)
b) sketch the graph of y=f(x), showing where it crosses the axes
c) write the polynomial in the form f(x)=ax^3+bx^2+cx+d (show all steps please)
thanks in advance

a)f(0)=> put 0 where there is an x
so (2(0)+1)(0+2)(0-3)=(1)(2)(-3)=-6

b) highest power of x is 3 so this takes shape of a cubic. where it cross the x axis is whenever the function equals 0 i.e.
(2x+1)(x+2)(x-3)=0
general known fact that when a function equals 0, put each bracket = 0 and solve so
2x+1=0 => x=-1/2
x+2=0 => x=-2
x-3=0 => x=3
so it takes the shape of x^3 crossing the x axis at -1/2, -2, and 3

c) expand first two brackets first
(2x+1)(x+2)= 2x^2 +4x+ x+ 2 = 2x^2 +5x +2
now do this times next bracket
(2x^2 +5x +2)(x-3)= 2x^3 -6x^2 +5x^2 -15x +2x -6 = 2x^3-x^2-13x-6

Hope I didn't make any mistakes.

a)(1)(2)(-3) = -6
b)crosses at x= -1/2, -2, and 3. It comes up on the left, wiggles at the x-axis thru it's 0's, then goes high on the right
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-/--\-|-/----- <---like that
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c)f(x) = (2x+1)(x+2)(x-3)
=(2x^2 + 4x + x + 2) (x-3) foil first 2 terms
=2x^3 - 6x^2 + 4x^2 - 24x + x^2 - 3x + 2x - 6 "foil" what's left
=2x^3 - x^2 -25x - 6 combine like terms