Question

Determine the equation of the following polynomials, using the given information. Write each polynomial in factored form and also in standard form. Don’t forget to solve for a!


A cubic whose zeros are at x = 2, x = -3 and x = 1 and the y-intercept is -12.

Answer 1

A polynomial with solution $x=x_0$ must be a multiple of $(x-x_0)$. So, your polynomial must be a multiple of

$(x-2)(x+3)(x-1)$

But this is already a cubic polynomial, so it can only be a multiple in a numerical sense:

$p(x)=a(x-2)(x+3)(x-1),\quad a\in\mathbb{R}$

The coefficient $a$ can be found imposing the condition about the y-intercept. The y-intercept is nothing more than the polynomial evaluated at zero:

$p(0)=a(-2)(3)(-1)=6a$

And we want this to be -12, so we have

$6a=-12\iff a=-2$

So, your polynomial is

$p(x)=-2(x-2)(x+3)(x-1)$

And if you expand it it becomes

$-2 x^3 + 14 x - 12$

Answer 2

Answer:

-2(x - 2)(x + 3)(x - 1) = 0.

-2x^3 + 14x - 12 = 0.

Step-by-step explanation:

In factored form it is:

f(x) = a(x - 2)(x + 3)(x - 1) = 0

Now, when  x = 0, y = -12 ( the y-intercept) so we have:

-12  = a(0-2)(0+3)(0-1)

6a = -12

a = -2.

The equation in factored form is

-2(x - 2)(x + 3)(x - 1) = 0.

In standard form:

-2(x - 2)(x + 3)(x - 1) = 0.

-2(x - 2)(x^2 + 2x - 3) = 0

-2( x^3 + 2x^2 - 3x - 2x^2 - 4x + 6) = 0

-2(x^3 - 7x + 6) = 0

-2x^3 + 14x - 12 = 0.