Question 1036226
<pre>
f(x) = x³ - 4x² + ax + b

tangent to the x-axis at x=3,

Therefore 3 is a double zero, (zero of even 
multiplicity, which can only be multiplicity 2,
Therefore synthetic division will produce a 
remainder of 0 twice:


  3|1 -4  a     b
   |<u>   3 -3   3a-9</u>  
    1 -1 a-3  3a+b-9 = 0


  3|1 -1 a-3
   |<u>   3  6</u>
    1  2 a+3 = 0

So f(x) factors as 

f(x) = (x-3)²(x+2)

So setting both those remainders = 0, gives
is a system of those two equations in two unknowns:

        a+3 = 0
          a = -3

      a+b-3 = 0
     -3+b-3 = 0
        b-6 = 0
          b = 6

f(x) = x³ - 4x² + ax + b
f(x) = x³ - 4x² - 3x + 18
 and its factorization is:

f(x) = (x-3)²(x+2)

The zeros are 3 and -2.  

The curve C and the x-as axis have points (3,0) and (-2,0)
in common. 

{{{graph(4000/23,400,-5,5,-3,20,x^3-4x^2-3x+18)}}}

To find the area bounded by C and the x-axis is given by

{{{A=int((x^3 - 4x^2 - 3x + 18),dx,-2,3)}}}
                       |3
   {{{expr(1/4)x^4-expr(4/3)x^3-expr(3/2)x^2+ 18x}}}| 
                       |-2

   {{{expr(1/4)3^4-expr(4/3)3^3-expr(3/2)3^2+ 18*3}}}{{{""-""}}}{{{expr(1/4)(-2)^4-expr(4/3)(-2)^3-expr(3/2)(-2)^2+ 18*(-2)}}}{{{""=""}}}{{{625/12}}}

Approximately 52.083

Edwin</pre>