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3 years ago

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Rewrite the expression without parentheses by using the Distributive Property p(q-r) VERY URGENT!

1 answer:

Vesnalui [34]3 years ago

4 0

Step-by-step explanation:

p(q-r) = pq - pr

Hope it helps

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

Svetlanka [38]

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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