Question 269216
Find a polynomial function f of degree 3 with leading coefficient 1 whose graph has x- intercepts at 
(-1,0), (2,0), (4,0)
<pre><font size = 4 color="indigo"><b> 
Two methods.  I'll do both:

======================================

Method 1:

Suppose the function is

{{{f(x)=x^3+Ax^2 + Bx + C}}}

then since it has x-intercept (-1,0) then when you substitute
-1 for x in that you get 0, so

{{{f(-1)=(-1)^3+A*(-1)^2+B*(-1)+C}}} which equals 0, so

{{{-1+A*(1)-B+C=0}}} or

{{{A-B+C=1}}}

and since it has x-intercept (2,0) then when you substitute
2 for x in that you get 0, so

{{{f(2)=(2)^3+A*(2)^2+B*(2)+C}}} which equals 0, so

{{{8+A*(4)+2B+C=0}}} or

{{{4A+2B+C=-8}}}

and since it has x-intercept (4,0) then when you substitute
4 for x in that you get 0, so

{{{f(4)=(4)^3+A*(4)^2+B*(4)+C}}} which equals 0, so

{{{64+A*(16)+4B+C=0}}} or

{{{16A+4B+C=-64}}}

So you have this system of 3 equations in 3 unknowns:

{{{system(A-B+C=1,4A+2B+C=-8,16A+4B+C=-64)}}}

Solve that and get A=-5, B=2 and C=8

So substitute in

{{{f(x)=x^3+Ax^2 + Bx + C}}}

and get

{{{f(x)=x^3-5x^2 + 2x + 8}}}

-----------------

Method 2:

(-1,0), (2,0), (4,0)

If you had solved {{{f(x)=0}}} you would have 
ended up with

    x = -1,       x = 2,       x = 4

before which you would have had

x + 1 =  0,   x - 2 = 0,   x - 4 = 0

before which you would have used the zero-factor
principle on this:

{{{(x + 1)(x - 2)(x - 4) = 0}}}

which you would have had after factoring, so before
factoring you would have had that multiplied out,
so we multiply it out

{{{(x^2-2x+x-2)(x-4)=0}}}

{{{(x^2-x-2)(x-4)=0}}}

{{{x^3-4x^2-x^2+4x-2x+8=0}}}

{{{x^3-5x^2+2x+8=0}}}

and since that was the solution to {{{f(x)=0}}}, then

{{{f(x)=x^3-5x+2x+8}}}

Same answer either way.  Take your pick.

Edwin</pre>