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OleMash [197]
3 years ago
9

Find a formula for the given polynomial.

Mathematics
2 answers:
levacccp [35]3 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]3 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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7 0
3 years ago
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Solve for x.<br><br> 2x^2 − 4x = 0
Schach [20]

Answer:

First, find a constant that can help you solve the equation. Note the equal sign, what you do to one side, you do to the other.

Remember, you are trying to do the following, changing: (note that y = constant)

x² - xy + y = (x - y)(x - y)

First, solve for the obvious answer. Plug in 0 to x:

2(0²) - 4(0) = 0

2(0) - 0 = 0

0 - 0 = 0

x = 0 is one of your answer choices.

Solve for the more complicated answer by doing what I put on the top:

2x² - 4x + 0 (+2) = 0 (+2)

2x² - 4x + 2 = 2

Divide 2 from both sides & all terms:

2(x² - 2x + 1) = 2(1)

x² - 2x + 1 = 1

Solve. Remember to set the equation = 0. Subtract 1 from both sides:

x² - 2x + 1 = 1

x² - 2x + 1 (-1) = 1 (-1)

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x² - 2x = 0

Isolate the -2x. Add 2x to both sides.

x² - 2x (+2x) = 0 (+2x)

x² = 2x

Isolate the variable x. Divide x from both sides.

(x²)/x = (2x)/x

x = 2x/x

x = 2

x = 2 is your other answer choice.

x = 0, 2 is your answer.

~

3 0
3 years ago
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