Question

What is the leading coefficient of a third degree polynomial function that has an output of 1,272 when x=2, and has zeros of −6, 7i, and −7i?

Answer 1

Answer:

The leading coefficient is 3

Step-by-step explanation:

Polynomials

Given the roots of a polynomial x1,x2,x3, it can be expressed as:

$p(x)=a(x-x1)(x-x2)(x-x3)$

Where a is the leading coefficient.

We are given the roots x1=-6, x2=7i, x3=-7i, thus:

$p(x)=a(x+6)(x-7i)(x+7i)$

Operating the product of the conjugated imaginary roots:

$p(x)=a(x+6)(x^2+49)$

Knowing p(2)=1,272 we can find the value of a

$p(2)=a(2+6)(4+49)=1,272$

Operating:

$a(8)(53)=1,272$

$424a=1,272$

Solving:

$a=1,272/424$

a=3

The leading coefficient is 3