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poizon [28]
3 years ago
5

Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.

Mathematics
1 answer:
Alisiya [41]3 years ago
3 0

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

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Answer:

\large \text{$x_1$}=-\dfrac{7}{4}, \large \text{$x_2$}=\dfrac{1}{2}

Step-by-step explanation:

Given: 8x² + 10x - 7 = 0

<u>When factoring trinomials of the form ax² + bx + c</u>:

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  • rewrite the polynomial with those factors replacing the middle term.

1. Split the middle term:

⇒ 8 × -7 = -56 [multiply the leading coefficient and the last term]

⇒ 14, -4 [find the product factors that add to the middle term's coefficient]

⇒ 8x² + 14x - 4x - 7 = 0 [rewrite with those factors replacing the middle term.]

2. Factor by grouping:

⇒ 8x² + 14x - 4x - 7 = 0 [factor out 2x]

⇒ 2x(4x + 7) - 4x - 7 = 0 [factor out -1 or the negative sign]

⇒ 2x(4x + 7) -1(4x + 7) = 0 [factor out 4x + 7]

⟹ (4x + 7)(2x - 1) = 0

3. Separate into 2 cases:

  1. 4x + 7 = 0
  2. 2x - 1 = 0

 

<u><em>Case 1:</em></u>

⇒ 4x + 7 = 0 [subtract 7 from both sides]

⇒ 4x + 7 - 7 = 0 - 7

⇒ 4x = -7 [divide both sides by 4]

⇒ 4x ÷ 4 = -7 ÷ 4

⟹ x = \bold{-\dfrac{7}{4}}

<u><em>Case 2:</em></u>

⇒ 2x - 1 = 0 [subtract 1 from both sides]

⇒ 2x - 1 + 1 = 0 + 1

⇒ 2x = 1 [divide both sides by 2]

⇒ 2x ÷ 2 = 1 ÷ 2

⟹ x = \bold{\dfrac{1}{2}}

Solutions:

\large \text{$x_1$}=-\dfrac{7}{4}, \large \text{$x_2$}=\dfrac{1}{2}

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

brainly.com/question/27638369

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Step-by-step explanation:

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Method 2:

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valentina_108 [34]

Answer:

A. there is a 99% probability that μ is between 3 and 9.

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From a random sample, we build a confidence interval, with a confidence level of x%.

The interpretation is that we are x% sure that the interval contains the true mean of the population.

In this problem:

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So between 6-3 = 3 and 6 + 3 = 9.

So we are 99% sure that the true population mean is between 3 and 9.

So the correct answer is:

A. there is a 99% probability that μ is between 3 and 9.

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(B) The difference between a particular solution and a general solution to an equation is that a particular solution is any specific figure that can satisfy the equation while a general solution is a statement that comprises all particular solutions of the equation.

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