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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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