How do you factor this polynomial?

Question

x^3 - 9x^2 + 27x - 27

I tried grouping the terms but I couldn't figure it out that way. Am I missing something?

by the way, you don't have to give me the answers...I'm looking for a method so I can do it myself.

Answer 1

Start with the constant term. What combination of three numbers when multiplied together give you -27? There are only a few. The most obvious ones are -3, -3, -3 and 3, 3, -3. (Of course there are other combinations of 27, 1, 1 and 9, 3, 1, but you'll quickly see that when you multiply these out, they don't work).

Since the x^3 has no coefficient, each term starts with x. So, now just try multiplying your combinations out.

You'll find that (x-3)(x-3)(x-3) or (x-3)^3 works just right.

Answer 2

x^3 - 9x^2 + 27x -27 = (x+a)(x^2+bx+c)
= x^3 + (a+b)x^2 + (ab + c)x + ac

so a+b = -9
ab + c = 27
ac = 27 . So assume an integral root.
then a must be +- 1 or 3 or 9 or 27
but if x < 0, then every term <0 so no negative root.

however 3 is a root. 27 - 81 + 81 - 27 = 0
therefore x-3 is a factor.

dividing x^3 - 9x^2 + 27x - 27 by x-3 gives
x^2 -6x + 9 = (x-3)^2

so x^3 - 9x^2 + 27x - 27 = (x-3)^3

Answer 3

Any zero must be a divisor of 27.
Also, if r is a zero, x-r is a factor.
We try 1, no good.
But 3 works,
Since 3³ -9*3² +27*3-27=0.
By synthetic division or long division, the quotient
is x² - 6x + 9 = (x-3)²
Thus x^3 - 9x^2 + 27x - 27 = (x-3)³.

Answer 4

This is just a simple application of a standard formula.

(a - b)^3 = a^3 - 3(a^2)b + 3a(b^2) - b^3

So,

x^3 - 9x^2 + 27x - 27 = (x - 3)^3

Answer 5

Try setting x to -2,-1,0,1, and 2. The majority of the time, x will be one of those numbers, thus leaving you with a regular quadratic equation.

Answer 6

It's a perfect cube. (x-3)^3

Answer 7

there are equations for cubing binomials...
(a+b)^3=a^3+3a^b+3ab^2+b^2
(a-b)^3=a^3-3a^b+3ab^2-b^2

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