Question

048x%20%2B%2040" id="TexFormula1" title=" {x}^{4} - 6 {x}^{3} + 22 {x}^{2} - 48x + 40" alt=" {x}^{4} - 6 {x}^{3} + 22 {x}^{2} - 48x + 40" align="absmiddle" class="latex-formula">
find the zeros​

Answer 1

Answer:

x = 2, 1 + 3i, 1 − 3i

Step-by-step explanation:

Find the Roots (Zeros)

x^4 − 6x^3 + 22x^2 − 48x + 40

Set x^4 − 6x^3 + 22x^2 − 48x + 40 equal to 0. x^4 − 6x^3 + 22x^2 − 48x + 40 = 0

Solve for x.

Factor the left side of the equation.

Factor x^4 − 6x^3 + 22x^2 − 48x + 40 using the rational roots test.

(x − 2) (x^3 − 4x^2 + 14x − 20) = 0

 Factor x^3 − 4x^2 + 14x − 20 using the rational roots test.

(x − 2) (x − 2) (x2 − 2x + 10) = 0

 Combine like factors.

(x − 2)2 (x^2 − 2x + 10) = 0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

(x − 2)^2 = 0

x^2 − 2x + 10 = 0

 Set (x − 2)^2 equal to 0 and solve for x.

Set (x − 2)^2 equal to 0.

 (x − 2)^2 = 0

Solve (x − 2)^2 = 0 for x.

x = 2

 Set x^2 − 2x + 10 equal to 0 and solve for x.

Set x^2 − 2x + 10 equal to 0. x^2 − 2x + 10 = 0

Solve x^2 − 2x + 10 = 0 for x.

Use the quadratic formula to find the solutions.

−b ± (√b^2 − 4 (ac) )/2a

Substitute the values a = 1, b = −2, and c = 10 into the quadratic formula and solve for x.

2 ± (√(−2)^2 − 4 ⋅ (1 ⋅ 10))/2 ⋅ 1

Simplify.

Simplify the numerator.

  x =    2 ± 6i/ 2.1

Multiply 2 by 1

 x =  2 ± 6i/2⋅1

 Simplify

  2 ± 6i/2  

   x = 1 ± 3i

The final answer is the combination of both solutions.

x = 1 + 3i, 1 − 3i

The final solution is all the values that make (x − 2)2 (x2 − 2x + 10) = 0 true.

x = 2, 1 + 3i, 1 − 3i