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Factor each completely.one zero has been given

f(x)=x^3-4x+16;-2

F(x) = X^4+ 2x^-13x^2 -10x + 40;2

F(x)=x^3-3x^2-13x+15;-3

F(x)=x^4-8x^3+19x^2-12x;3

Sagot :

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The polynomials in factored form are listed below:

  1. f(x) = (x - 3.043) · (x - 1.521 - i 1.716) · (x - 1.521 + i 1.716)
  2. f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)
  3. f(x) = (x + 3) · (x - 5) · (x - 1)
  4. f(x) = (x - 4) · (x - 3) · (x - 1) · x

How to factor polynomials

In this question we must factor polynomials as there are both numerical and analytical methods. Mathematically speaking, factoring polynomials is represented by the following formula:

[tex]\sum^{n}_{i = 0} \,c_{i}\cdot x^{i} = \prod^{n}_{i=1}(x-r_{i})[/tex], [tex]\forall\,x, r_{i}\in \mathbb{C}[/tex]   (1)

Where:

  • [tex]c_{i}[/tex] - [tex]i[/tex]-th Coefficient
  • [tex]r_{i}[/tex] - [tex]i[/tex]-th Root

Regarding fourth order polynomials we can solve them by Ferrari's method and third order polynomials by Descartes' method. Then, the solutions of each polynomials are given below:

Polynomial 1 ([tex]f(x) = x^{3}-4\cdot x +16[/tex])

f(x) = (x - 4) · (x - 3) · (x - 1) · x

Polynomial 2 ([tex]f(x) = x^{4}+2\cdot x^{3}-13\cdot x^{2}-10\cdot x + 40[/tex])

f(x) = (x - 2.236) · (x - 2) · (x + 2.236) · (x + 4)

Polynomial 3 ([tex]f(x) = x^{3}-3\cdot x^{2}-13\cdot x +15[/tex])

f(x) = (x + 3) · (x - 5) · (x - 1)

Polynomial 4 ([tex]f(x) = x^{4}-8\cdot x^{3}+19\cdot x^{2}-12\cdot x[/tex])

f(x) = (x - 4) · (x - 3) · (x - 1) · x

To learn more on polynomials, we kindly invite to check this verified question: https://brainly.com/question/17822016

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