Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the
leading coefficient is 1 .

Question

Find a polynomial function of degree 3 with the given numbers as zeros. Assume that the leading coefficient is 1 . $$ -2 ; 6,8 $$

UpStudy AI · Accepted Answer

Consider the zeros $$ -2 $$, $$ 6 $$, and $$ 8 $$. Since the polynomial has leading coefficient 1, it can be written in factored form as

$$
P(x) = (x - (-2))(x - 6)(x - 8) = (x + 2)(x - 6)(x - 8).
$$

Now, multiply the factors step by step.

First, multiply the factors $$(x - 6)$$ and $$(x - 8)$$:

$$
(x - 6)(x - 8) = x^2 - 8x - 6x + 48 = x^2 - 14x + 48.
$$

Next, multiply this result by $$(x + 2)$$:

$$
P(x) = (x + 2)(x^2 - 14x + 48).
$$

Distribute $$(x + 2)$$ into the quadratic expression:

$$
P(x) = x(x^2 - 14x + 48) + 2(x^2 - 14x + 48).
$$

Compute each product:

$$
x(x^2 - 14x + 48) = x^3 - 14x^2 + 48x,
$$
$$
2(x^2 - 14x + 48) = 2x^2 - 28x + 96.
$$

Now, combine like terms:

$$
P(x) = x^3 - 14x^2 + 48x + 2x^2 - 28x + 96.
$$

$$
P(x) = x^3 + (-14x^2 + 2x^2) + (48x - 28x) + 96,
$$

$$
P(x) = x^3 - 12x^2 + 20x + 96.
$$

Thus, the polynomial function is

$$
P(x) = x^3 - 12x^2 + 20x + 96.
$$
The polynomial function is $$ P(x) = x^3 - 12x^2 + 20x + 96 $$.

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