Question

write a linear factorization of x^3+4x-5

Answer 1

Possible roots = +/-1, +/-2, +/-5

Let's test if 2 is a possible zero.

8 - 24 + 26 - 10

-2 + 2 = 0

2 is one of the zeroes.

Use synthetic division and divide the expression by (x-2). Then multiply that by (x-2).

(x^3 - 6x^2 + 13x - 10) / (x-2) = x^2 - 4x + 5

(x-2)(x^2-4x+5)

Use quadratic formula to find the zeros found in \(x^2-4x+5\)

x = 2+/-i

Zeroes: 2, 2+/-i

Answer 2

But two doesn't work for that

Answer 3

The only real zero should be 1