What's the real solution of x^4-6x^3+22x+15

Answer 1

By trial and error, 3 is a root, so x - 3 is a factor.

Divide out by x - 3 to get x⁴ - 6x³ + 22x + 15 = (x - 3)(x³ - 3x² - 9x - 5)

By trial and error, -1 is a root of x³ - 3x² - 9x - 5, so x + 1 is a factor.

Divide out by x + 1 to get (x - 3)(x + 1)( x² - 4x - 5) = (x - 3)(x + 1)(x + 1)(x - 5) = (x - 3)(x - 5)(x + 1)²

The real roots are 3, 5, and -1 (-1 is a root of multiplicity 2).

Answer 2

Factors of 15: ±(1,3,5)

Use synthetic division, trying -1

1 -6 0 22 15 |-1

-1 7 -7 -15

---------------------

1 -7 7 15 0 |-1

1 -7 7 15 |-1 Try -1 again

-1 8 -15

--------------------

1 -8 15 0

x²-8x+15

Factors into (x-3)(x-5)

(x+1)(x+1)(x-3)(x-5)=0

x = -1,-1,3,5