wickerh23
wickerh23 wickerh23
  • 18-01-2020
  • Mathematics
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Algebra 2

Please see the image below for my question!

Algebra 2 Please see the image below for my question class=

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kudzordzifrancis
kudzordzifrancis kudzordzifrancis
  • 30-01-2020

Answer:

[tex]x=\pm i,x=-13[/tex]

Step-by-step explanation:

Since [tex]x=-1[/tex] is a solution of [tex]x^4+14x^3+14x^2+14x+13[/tex], by the factor theorem, [tex](x+1)[/tex] is a factor.

We perform the long division as shown in the attachment to obtain:

[tex]x^4+14x^3+14x^2+14x+13=(x+1)(x^3+13x+x+13)[/tex]

We now factor the cubic polynomial by grouping.

[tex]x^4+14x^3+14x^2+14x+13=(x+1)(x^2(x+13)+1(x+13))[/tex]

[tex]x^4+14x^3+14x^2+14x+13=(x+1)(x^2+1)(x+13)[/tex]

Hence the other roots are:

[tex]x=\pm i,x=-13[/tex]

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