Answered

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3. What are the roots of the polynomial y = x³ - 8?

Sagot :

culveryolanda46

Step-by-step explanation:

x^3 is a perfect cube, 8 is a perfect cube, so we use difference of cubes.

[tex] {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )[/tex]

Cube root of x^3 is x.

Cube root of 8 is 2

So

a=x

b= 2.

[tex](x - 2)( {x}^{2} - 2x + 4)[/tex]

Set these equations equal to zero

[tex]x - 2 = 0[/tex]

[tex]x = 2[/tex]

[tex] {x}^{2} - 2x + 4 = 0[/tex]

If we do the discriminant, we get a negative answer so we would have two imaginary solutions,

Thus the only real root is 2.

If you want imaginary solutions, apply the quadratic formula.

[tex]1 + i \sqrt{ 3 } [/tex]

and

[tex]1 - i \sqrt{3} [/tex]

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