# f(x)=x^5-4x^3-x^2+4?

the real factors are (x+2), (x-1), and (x-2)

how do I find the imaginary factors using long polynomial division

### 3 Answers

- Φ² = Φ+1Lv 72 months ago
(x+2)(x-1)(x-2) = x^3 - x^2 - 4x + 4

From the first and last term the quadratic yielding the complex solutions is x^2 + ax + 1.

(x^3 - x^2 - 4x + 4)(x^2 + ax + 1) = x^5 + (a-1)x^4 + .... + 4. a - 1 = 0, so the quadratic is x^2 + x + 1, which factors to (x + ½ + ½√3 i)(x + ½ - ½√3 i).

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- ted sLv 72 months ago
either multiply those factors together , then divide f by that result...OR divide f by ' x + 2 ' , that result by ' x - 1 ' , that result by x - 2...to get a quadratic function....use quadratic equation to find the last 2 zeros...( x - r_1) (x - r_2)

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- JimLv 72 months ago
Since you have a quadratic left, I would simply use Quadratic formula or Completing the squares.

(x^2 + x + 1) = 0

Subtract 1 from both sides: x^2 + x = -1

Add 1/4 to both sides: x^2 + x + 1/4 = -3/4

Write the left hand side as a square: (x + 1/2)^2 = -3/4

Take the square root of both sides: x + 1/2 = (i sqrt(3))/2 or x + 1/2 = -(i sqrt(3))/2

Subtract 1/2 from both sides: x = (i sqrt(3))/2 - 1/2 or x + 1/2 = -(i sqrt(3))/2

Subtract 1/2 from both sides:

x = (i sqrt(3))/2 - 1/2 or x = -(i sqrt(3))/2 - 1/2

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