Respuesta :

altavistard

Answer:

3 ± i5

Step-by-step explanation:

Here we're given four sets of possible roots of the given polynomial.   Each set consists of two complex quantities and 1 real quantity.

First, we determine whether +4 is a root, then whether -4 is a root.  Let's use synthetic division to do that:

     -----------------------

4   /  1   -2    10   136

             4      8    72

    ---------------------------

        1     2    18    208     Since the remainder is not zero, 4 is not a root.

Eliminate the first two possible answer choices, and assume that -4 is a root.

Let's check this out to be certain:

     -----------------------

-4  /  1   -2    10   136

             -4   24  -136

    ---------------------------

        1     -6    34    0

Since the rem. is zero, -4 is a root, and the coefficients of the 2nd-degree quotient are 1, -6 and 34.

In other words, a = 1, b = -6 and c = 34.

Let's apply the quadratic rule to find the roots:

       6 ± √(36 - 4[1][34] )       6 ± √ (-100)

x = ------------------------------ = ----------------------- = 3 ± i5

                     2                                   2

So the correct answer is the last one of the four given possible answers:

3 ± i5

Answer:

[tex]3+5i, 3-5i,-4[/tex]

Step-by-step explanation:

Given is a cubic equation in x as

[tex]x^3-2x^2+10x+136=0[/tex]

We know by remainder theorem if f(a) =0 then x=a is a root

Constant term = 136 has factors as 1,2,4,8, 17, 34

We find that f(a) not 0 for a =1,2.

[tex]f(1) = 144\\f(2) = 156\\f(-1) = 123\\f(-4) = -64-2(16)+10(-4)+136 = 0[/tex]

Hence [tex]x=-4[/tex] is one zero.

We can divide by x+4 to make it as a quadratic equation

[tex]x^3-2x^2+10x+136=(x+4)(x^2-6x+34)[/tex]

Now using quadratic formula we can find other roots

[tex]x^2-6x+34=0[/tex] has roots as

[tex]x=\frac{6+/-\sqrt{36-136} }{2} \\=3+5i, 3-5i[/tex]

Thus roots are

[tex]3+5i, 3-5i,-4[/tex]

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