How do you factor x^3-9x^2+25x-21, given that x=3 is a zero ?

1 Answer
George C.
Nov 2, 2016

x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))

Explanation:

h(x) = x^3-9x^2+25x-21

The difference of squares identity can be written:

a^2-b^2 = (a-b)(a+b)

We use this later with a = (x-3) and b=sqrt(2)

We are told that 3 is a zero, so (x-3) must be a factor:

x^3-9x^2+25x-21 = (x-3)(x^2-6x+7)

Then, completing the square we find:

x^2-6x+7 = x^2-6x+9-2

color(white)(x^2-6x+7) = (x-3)^2-(sqrt(2))^2

color(white)(x^2-6x+7) = ((x-3)-sqrt(2))((x-3)+sqrt(2))

color(white)(x^2-6x+7) = (x-3-sqrt(2))(x-3+sqrt(2))

Putting it all together:

x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))

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