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Varvara68 [4.7K]
2 years ago
11

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?
Mathematics
1 answer:
Gre4nikov [31]2 years ago
6 0

Answer:

<em>The leading coefficient is 3</em>

Step-by-step explanation:

<u>Polynomials</u>

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

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