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statuscvo [17]
3 years ago
13

Write the simplest polynomial function for each set of zeros

Mathematics
1 answer:
dexar [7]3 years ago
5 0

9514 1404 393

Answer:

  p(x) = x^4 -5x^3 +20x -16

Step-by-step explanation:

If 'a' is a zero of the polynomial, then (x -a) will be a factor. For the given zeros, the simplest polynomial will be the product of the corresponding factors:

  p(x) = (x -2)(x +2)(x -4)(x -1) . . . . . . note that x -(-2) = x +2 (factored form)

__

Multiplying these out gives the result in standard form.

The product of the <em>first two factors</em> is a "special product" recognizable as the difference of two squares.

  (x -2)(x +2) = x^2 -2^2 = x^2 -4

The product of the <em>last two factors</em> can be found in the usual way. The distributive property applies.

  (x -4)(x -1) = x(x -1) -4(x -1)

  = x^2 -x -4x +4 = x^2 -5x +4

Then the full polynomial is the product of these partial products:

  p(x) = (x^2 -4)(x^2 -5x +4)

  = x^2(x^2 -5x +4) -4(x^2 -5x +4)

  = x^4 -5x^3 +4x^2 -4x^2 +20x -16

  p(x) = x^4 -5x^3 +20x -16 . . . . . . . . standard form

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Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

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Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4×(1/2)^{\infty}

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

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BlackZzzverrR [31]

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