Find polynomial function?

Question

Find a polynomial function​ P(x) of degree 3 with real coefficients that satisfies the given conditions. Do not use a calculator.

Zeros of −7​, 1​, and​ 0; P(-2)= -10

Answer 1

We know that from factoring that a polynomial completely factored will reveal the zeros of that function. For us, we have x = -7, 1, 0.

This means that we have a function that looks like: y = a( x(x+7)(x-1) ).

Our a constant is a stretching constant which will be calculated from our point. The reason we do this is the "stretch" of function to fit where we want (in our case the point (-2, -10)) without changing the zeros of our function.

We have y = a( x(x+7)(x-1) )

= a( x(x² + 6x - 7) )

= a( x³ + 6x² - 7x )

Now we want to find a. We have a value for x and y from our point. x = -2, y = -10

Substituting:

-10 = a( (-2)³ + 6(-2)² - 7(-2) )

-10 = a( -8 + 24 + 14)

-10 = a(30)

-10/30 = a

a = -1/3

Now we can put our value for a into our equation.

Our final equation is:

y = -1/3( x³ + 6x² - 7x )

= -1/3•x³ - 2•x² + 7/3•x

Answer 2

P(x) = a(x + 7)(x - 1)(x - 0)

-10 = a(2 + 7)(2 - 1)(2)

-10 = 18a

-5 / 9 = a