Question

Find the roots of the polynomial equation.

Answer 1

3 ± i5

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

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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

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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 2

`3+5i, 3-5i,-4`

Explanation:

Given is a cubic equation in x as

`x^3-2x^2+10x+136=0`

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.

`f(1) = 144\\f(2) = 156\\f(-1) = 123\\f(-4) = -64-2(16)+10(-4)+136 = 0`

Hence
`x=-4` is one zero.

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

`x^3-2x^2+10x+136=(x+4)(x^2-6x+34)`

Now using quadratic formula we can find other roots

`x^2-6x+34=0` has roots as

`x=(6+/-√(36-136) )/(2) \\=3+5i, 3-5i`

Thus roots are

`3+5i, 3-5i,-4`