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lubasha [3.4K]
3 years ago
5

Part A: Create a third-degree polynomial in standard form. How do you know it is in standard form? (5 points)

Mathematics
1 answer:
Svetlanka [38]3 years ago
4 0

Answer:

(See explanation for further details)

Step-by-step explanation:

a) Let consider the polynomial p(x) = 5\cdot x^{3} +2\cdot x^{2} - 6 \cdot x +17. The polynomial is in standards when has the form p(x) = \Sigma \limit_{i=0}^{n} \,a_{i}\cdot x^{i}, where n is the order of the polynomial. The example has the following information:

n = 3, a_{0} = 17, a_{1} = -6, a_{2} = 2, a_{3} = 5.

b) The closure property means that polynomials must be closed with respect to addition and multiplication, which is demonstrated hereafter:

Closure with respect to addition:

Let consider polynomials p_{1} and p_{2} such that:

p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i} and p_{2} = \Sigma \limits_{i=0}^{n}\,b_{i}\cdot x^{i}, where m \geq n

p_{1}+p_{2} = \Sigma \limits_{i=0}^{n}\,(a_{i}+b_{i})\cdot x^{i} + \Sigma_{i=n+1}^{m}\,a_{i} \cdot x^{i}

Hence, polynomials are closed with respect to addition.

Closure with respect to multiplication:

Let be p_{1} a polynomial such that:

p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}

And \alpha an scalar. If the polynomial is multiplied by the scalar number, then:

\alpha \cdot p_{1} = \alpha \cdot \Sigma \limits_{i = 0}^{m}\,a_{i}\cdot x^{i}

Lastly, the following expression is constructed by distributive property:

\alpha \cdot p_{1} = \Sigma \limits_{i=0}^{m}\,(\alpha\cdot a_{i})\cdot x^{i}

Hence, polynomials are closed with respect to multiplication.

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If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?
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\boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given \boxed{ \ a(x) = 3x + 1 \ } and \boxed{ \ b(x) = \sqrt{x - 4} \ }.

We will form (b o a)(x) and then determine the domain.

<u>Step-1</u>

\boxed{ \ (b \circ a)(x) = b(a(x)) \ }

Replace each appearance of x in b(x) with \boxed{ \ a(x) = 3x + 1 \ }.

\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }

Thus, \boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }

<u>Step-2</u>

To be defined, the value under the radical sign must not be negative. Therefore, the domain of (b \circ a)(x) = \sqrt{3x - 3} are processed as follows.

\boxed{ \ 3x - 3 \geq 0 \ }

Both sides added by 3.

\boxed{ \ 3x \geq 3 \ }

Both sides divided by 3.

\boxed{ \ x\geq 1 \ }

Thus, the domain of (b \circ a)(x) = \sqrt{3x - 3} is \boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Learn more</h3>
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Answer: R would equal -3

Step-by-step explanation:

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