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

Answer 1

3, 5, -1, -1, four real roots

by decartes' sign rule

there are either 0 or 2 + real roots

and 0 or 2 negative real roots

0,2 or 4 imaginary roots

total 4 roots = to the degree of the polynomial

x^4 -6x^3 +22x +15, try integer factors of 15, such as + or - 1, 3, 5, 15

x f(x)

0 15

1 1- 6+ 22 +15 = 32

2 16 - 48 +44 +15 = 27

3 81 - 162 + 66 +15 = 0 Bingo x= 3 is one root

-1 1+5 -22 +16=0 Bingo another root

etc

just keep trying more single digit integers

or just graph the polynomial and locste the x intercepts = the real roots

x=3 and 5 are two. x=-1 will be a point of tangency, indicating multiplicyt of two

the polynomial factored = (x-3)(x-5)(x+1)^2

Answer 2

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 3

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