Answer:
h(x) = (x -1)(x -(3 +√2))(x -(3 -√2))
Explanation:
The given cubic is said to have 1 as a zero. That means (x -1) is a linear factor of the equation. Dividing the cubic by that using your favorite method gives a quotient that is the quadratic ...
x^2 -6x +7
We can write this in vertex form by "completing the square":
(x^2 -6x) +7
= (x^2 -6x +9) +7 -9 . . . . . add and subtract (-6/2)^2 = 9
= (x -3)^2 -2
This can be factored as the difference of two squares:
a² -b² = (a -b)(a +b)
Here we have ...
a² = (x-3)² ⇒ a = (x -3)
b² = 2 ⇒ b = √2
Then the linear factors of the quadratic are ...
x^2 -6x +7 = ((x -3) -√)((x -3) +√2)
= (x -(3 +√2))(x -(3 -√2)) . . . . . using the associative property of addition
And the cubic can be factored to linear factors as ...
h(x) = x^3 -7x^2 +13x -7
h(x) = (x -1)(x^2 -6x +7)
h(x) = (x -1)(x -(3 +√2))(x -(3 -√2))