SOLUTION: For the​ following, find the function P defined by a polynomial of degree 3 with real coefficients that satisfy the given conditions. The zeros are​ -3,-​1, and 4. p(-2)=-1

Algebra ->  Equations -> SOLUTION: For the​ following, find the function P defined by a polynomial of degree 3 with real coefficients that satisfy the given conditions. The zeros are​ -3,-​1, and 4. p(-2)=-1      Log On


   

Question 1192629: For the​ following, find the function P defined by a polynomial of degree 3 with real coefficients that satisfy the given conditions.
The zeros are​ -3,-​1, and 4.
p(-2)=-18
p(x)=
thanks your your help!(:
Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

The zeros are​ x%5B1%5D=-3, x%5B2%5D=-1, and x%5B3%5D=4

p%28-2%29=-18

p%28x%29=a%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29.....substitute zeros
p%28x%29=a%28x-%28-3%29%29%28x-%28-1%29%29%28x-4%29
p%28x%29=a%28x%2B3%29%28x%2B1%29%28x-4%29
p%28x%29=a%28x%2B3%29%28x%2B1%29%28x-4%29.....to find a use given p%28-2%29=-18
-18=a%28-2%2B3%29%28-2%2B1%29%28-2-4%29
-18=a%281%29%28-1%29%28-6%29
-18=6a
a=-18%2F6
a=-3
substitute in p%28x%29=a%28x%2B3%29%28x%2B1%29%28x-4%29
p%28x%29=-3%28x%2B3%29%28x%2B1%29%28x-4%29.........expand
p%28x%29=-3x%5E3+%2B+39x+%2B+36

+graph%28+600%2C+600%2C+-10%2C+10%2C+-30%2C+30%2C+-3x%5E3%2B39x%2B36%29+









Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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: P(x) = -3x^3 + 39x + 36