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Need on these two cause i’m very lost on them

Posted on October 23, 2021 By Spooderfaxy7813 3 Comments on Need on these two cause i’m very lost on them

Need on these two cause i’m very lost on them


[tex]Need on these two cause i’m very lost on them[/tex]

Mathematics

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Comments (3) on “Need on these two cause i’m very lost on them”

  1. Expert says:
    October 23, 2021 at 6:18 pm

    Dis the correct answer

    Reply
  2. Expert says:
    October 23, 2021 at 9:21 pm

    215.92 dollars is the answer

    Reply
  3. sarahsompayrac says:
    October 24, 2021 at 1:02 am

    see explanation

    Step-by-step explanation:

    Given

    f(x) = (x - 4)(2x - 1)²(x - 2)²

    To find the roots equate f(x) to zero, that is

    (x - 4)(2x- 1)²(x - 2)² = 0

    Equate each of the factors to zero and solve for x

    x - 4 = 0 ⇒ x = 4

    2x - 1 = 0 ⇒ x = [tex]\frac{1}{2}[/tex]  ← with multiplicity 2

    x - 2 = 0 ⇒ x = 2 ← with multiplicity 2

    Hence the roots are

    { 4, [tex]\frac{1}{2}[/tex], 2 }

    Given

    f(x) = x³ + 4x² + 7x + 6

    Note that

    f(- 2) = (- 2)³ + 4(- 2)² + 7(- 2) + 6 = - 8 + 16 - 14 + 6 = 0

    Since f(- 2) = 0 then by the factor theorem x = - 2 is a root and (x + 2) a factor

    Using synthetic division

    - 2 | 1   4    7    6

             - 2  - 4  - 6

    --------------

         1    2    3    0

    Thus

    f(x) = (x + 2)(x² + 2x + 3)

    Solve x² + 2x + 3 using the quadratic formula

    with a = 1, b = 2 and c = 3

    x = (- 2 ± [tex]\sqrt{2^2-(4(1)(3)}[/tex] ) / 2

      = ( - 2 ± [tex]\sqrt{4-12}[/tex] ) / 2

      = ( - 2 ± [tex]\sqrt{-8}[/tex] ) / 2

      = (- 2 ± 2i[tex]\sqrt{2}[/tex] ) / 2

      = - 1 ± i[tex]\sqrt{2}[/tex]

    Hence roots are

    { - 2, - 1 + i[tex]\sqrt{2}[/tex], - 1 - i[tex]\sqrt{2}[/tex] }

         

    Reply

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