3 years ago

Identify the variables, the constant, the coefficients, and the exponents in the algebraic

Mathematics

1 answer:

Naya [18.7K]3 years ago

4 0

Variables: x and y
Coefficients: 5 and 2
Constants: 4
Exponents: 2 (if 3y is being squared)

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P(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).

Mila [183]

Answer:

The polynomial is:

$p(x) = -x^3 - 2x^2 + 5x + 6$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$ , in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that $x_1 = -3, x_2 = -1, x_3 = 2$ . Thus

$p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)$

$p(x) = a(x - (-3))*(x - (-1))*(x-2)$

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

$p(x) = a(x^2+4x+3)(x-2)$

$p(x) = a(x^3+2x^2-5x-6)$

Passes through the point (1,12).

This means that when $x = 1, p(x) = 12$ . We use this to find a.

$12 = a(1 + 2 - 5 - 6)$

$-12a = 12$

$a = -\frac{12}{12}$

$a = -1$

Thus

$p(x) = -(x^3+2x^2-5x-6)$

$p(x) = -x^3 - 2x^2 + 5x + 6$

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