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OleMash [197]
2 years ago
9

Find a formula for the given polynomial.

Mathematics
2 answers:
levacccp [35]2 years ago
4 0

In this question, we have to identify the zeros of the polynomial, along with a point, and then we get that the formula for the polynomial is:

p(x) = -0.5(x^3 - x^2 + 6x)

------------------------

Equation of a polynomial, according to it's zeros:

Given a polynomial f(x), this polynomial has roots such that it can be written as: , in which a is the leading coefficient.

------------------------

Identifying the zeros:

Given the graph, the zeros are the points where the graph crosses the x-axis. In this question, they are:

x_1 = -2, x_2 = 0, x_3 = 3

Thus

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

p(x) = a(x - (-2))(x - 0)(x-3)

p(x) = ax(x+2)(x-3)

p(x) = ax(x^2 - x + 6)

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

------------------------

Leading coefficient:

Passes through point (2,-8), that is, when x = 2, y = -8, which is used to find a. So

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

-8 = a(2^3 - 2^2 + 6*2)

16a = -8

a = -\frac{8}{16} = -0.5

------------------------

Considering the zeros and the leading coefficient, the formula is:

p(x) = -0.5(x^3 - x^2 + 6x)

A similar problem is found at brainly.com/question/16078990

max2010maxim [7]2 years ago
4 0

The formula that represents the polynomial in the figure is p(x) = x^{3}-x^{2}-6\cdot x.

Based on the Fundamental Theorem of Algebra, we understand that Polynomials with real Coefficient have <em>at least</em> one real Root and <em>at most</em> a number of Roots equal to its Grade. The Grade is the maximum exponent that Polynomial has and root is a point such that p(x) = 0. By Algebra we understand that polynomial can be represented in this manner known as Factorized form:

p(x) = \Pi\limits_{i=0}^{n} (x-r_i) (1)

Where:

n - Grade of the polynomial.

i - Index of the root binomial.

x - Independent variable.

We notice that polynomials has three roots in x = -2, x = 0 and x = 3, having the following construction:

p(x) =(x+2)\cdot x \cdot (x-3)

p(x) = (x^{2}+2\cdot x)\cdot (x-3)

p(x) = x^{3}+2\cdot x^{2}-3\cdot x^{2}-6\cdot x

p(x) = x^{3}-x^{2}-6\cdot x

The formula that represents the polynomial in the figure is p(x) = x^{3}-x^{2}-6\cdot x.

Here is a question related to the determination polynomials: brainly.com/question/10241002

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Answer in fraction form: \fbox {-10/39}
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