Question 1192629
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The term "root" is another way of saying the zeros of a polynomial. 
These are the x values that make y = P(x) = 0.
Visually they are the x intercepts where the curve crosses or touches the x axis.


The roots are: -3, -1, 4


This means
x = -3 or x = -1 or x = 4
turns into
x+3 = 0 or x+1 = 0 or x-4 = 0


Use the Zero Product Property to connect those three equations into one single equation.
(x+3)(x+1)(x-4) = 0


Let's expand out the (x+1)(x-4) portion
(x+1)(x-4)
y(x-4) .... let y = x+1
xy-4y .... distributive property
x(y) - 4(y)
x(x+1) - 4(x+1) ... plug in y = x+1
x^2+x - 4x - 4  .... distributive property
x^2 - 3x - 4


The expression (x+1)(x-4) expands out to x^2-3x-4
You can use the FOIL rule as an alternative path


So
(x+3)(x+1)(x-4) = 0
becomes
(x+3)(x^2-3x-4) = 0


Furthermore,
(x+3)(x^2-3x-4) = 0
w(x^2-3x-4) = 0 ..... let w = x+3
wx^2 - 3wx - 4w = 0  .... distributive property
x^2( w ) - 3x( w ) - 4( w ) = 0
x^2( x+3 ) - 3x( x+3 ) - 4( x+3 ) = 0 ... plug in w = x+3
x^3 + 3x^2 - 3x^2 - 9x - 4x - 12 = 0  .... distributive property
x^3 - 13x - 12 = 0


So the claim is that
P(x) = x^3 - 13x - 12


Let's see what happens when we plug in x = -2
P(x) = x^3 - 13x - 12
P(-2) = (-2)^3 - 13(-2) - 12
P(-2) = 6
We should get to -18 instead of 6


To fix this, multiply by -3
6*(-3) = -18


This means we need a -3 out front like this
P(x) = -3(x^3 - 13x - 12)
which distributes to get
P(x) = -3x^3 + 39x + 36


To verify the answer, plug in the following x values:
x = -3, x = -1, x = 4, x = -2
And you should get these y values in the exact order presented
y = 0, y = 0, y = 0, y = -18
I'll leave the verification steps for you to do.


Answer: <font color=red>P(x) = -3x^3 + 39x + 36</font>
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