Answer:
h(x) = (x -1)(x +3+√2)(x +3-√2)
Step-by-step explanation:
You want the linear factors of h(x) = x^3 +5x^2 +x -7, given that 1 is a zero.
Quotient
If 1 is a zero, then (x -1) is a factor. The remaining quadratic factor can be found using synthetic division, as shown in the attachment. This tells us the factorization is ...
h(x) = (x -1)(x^2 +6x +7)
Quadratic factors
The factors of the quadratic can be found by completing the square:
x^2 +6x = -7
x^2 +6x +9 = 2 . . . . . . . add (6/2)^2 = 9 to both sides
(x +3)^2 = 2 . . . . . . . . show as a squre
x +3 = ±√2 . . . . . . . . take the square root
x +3 ±√2 = 0 . . . . . . subtract ±√2; the two remaining factors
This tells us that h(x) = 0 when (x +3 +√2) = 0 and when (x +3 -√2) = 0. These binomial expressions are the remaining linear factors of h(x).
h(x) = (x -1)(x +3+√2)(x +3-√2)
