solve x^5-12x^4+54x^3-108x^2+81x=0

Question

Pre-calculus

Answer 1

Part 2

Now we need to factor x^3 - 9x^2 + 27x - 27.  It's the same possible solutions (x= +-3, +-9, +-27, +-81), but we've already shown that the only one of these solutions that works is x = 3.  Let's see if we can factor (x-3) out again:

				x^2	-6x	+9
x	-3		x^3	-9x^2	+27x	-27
			-(x^3	-3x^2)
				-6x^2	+27x
				-(-6x^2	+18x)
					9x	-27
					-(9x	-27)
						0
		x

That tells us that x^3 - 9x^2 +27x - 27 = (x-3)(x^2 - 6x + 9)

Now that we're down to x^2 - 6x + 9, it factors to (x-3)(x-3).

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

And:  x^4 - 12x^3 +54x^2 - 108x + 81 = (x-3)(x-3)(x-3)(x-3)

And and:  x^5 - 12x^4 + 54x^3 - 108x^2 + 81x = x(x-3)(x-3)(x-3)(x-3)

Our original problem now looks like this:

x(x-3)(x-3)(x-3)(x-3) = 0

Solutions: x = 0, 3, 3, 3, 3 or just x = 0, 3

Answer:  x = 0, 3

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Note:  I gave myself a bonus of -40 points to counteract answering this question in 2 parts (+40 points).

Answer 2

Part 1

x^5 - 12x^4 + 54x^3 - 108x^2 + 81x = 0

x(x^4 - 12x^3 + 54x^2 - 108x + 81) = 0

If there's a nice (whole number solution) way to factor x^4 - 12x^3 + 54x^2 - 108x + 81, it has to be made of factors of 81 (3*3*3*3).

Think about a solution (x = something) that isn't made of 3's.  Let's say, x = 5.

No matter what we do, the part with x's (x^4 - 12x^3 + 54x^2 - 108x) will result in a number divisible by 5.  But the whole thing (x^4 - 12x^3 + 54x^2 - 108x + 81) has to be 0, so the part with the x's has to equal -81.  -81 is not divisible by 5, so this doesn't work.  x = 5 is not a solution.

Possible solutions to x^4 - 12x^3 + 54x^2 - 108x + 81 = 0 are  x= +-3, +-9, +-27, +-81

f(x) = x^4 - 12x^3 + 54x^2 - 108x + 81.  We want to find x = ? so that f(?) = 0

f(3) = 0

f(-3) = 1296

f(9) = 1296

f(-9) = 20736

f(27) = 331776

f(-27) = 810000

f(81) = 37015056

f(-81) = 49787136

It looks like the only nice whole number solution is x = 3, which means we want to factor x-3 out from x^4 - 12x^3 + 54x^2 - 108x + 81.  It's a visit to our friend, Mr. Synthetic Division.

				x^3	-9x^2	+27x	-27
x	-3		x^4	-12x^3	+54x^2	-108x	+81
			-(x^4	-3x^3)
				-9x^3	+54x^2
				-(-9x^3	+27x^2)
					27x^2	-108x
					-(27x^2	-81x)
						-27x	+81
						-(-27x	+81)
		x					0

x^4 - 12x^3 +54x^2 - 108x + 81 = (x-3)(x^3 - 9x^2 +27x - 27)

Answer 3

Given: f(x)=x^5-12x^4+54x^3-108x^2+81x=0 ···Eq.1

Let f(x)=xg(x)=0, so x=0 or g(x)=0 and g(x)=x^4-12x^3+54x^2-108x+81   Here, we examine g(x)=0.

Notice that the coefficients of each term in g(x) are 1,-12,54,-108,81, so the greatest common factor of 54,-108, 81 is 27, the sum of 2 factors of 27, 3 and 9, is 12, and the product of 3 and 27 is 81.   Thus, We divide g(x) into 2 parts assuming (x-3) might be one of the factors, getting

g(x)=(x^4-12x^3+27x^2)+(27x^2-108x+81)=x^2(x^2-12x+27)+27(x^2-4x+3)

=x^2(x-3)(x-9)+27(x-3)(x-1)=(x-3){x^2(x-9)+27(x-1)}=(x-3)(x^3-9x^2+27x-27)

Apply the fomula: (a-b)^3=a^3-3(a^2)b+3ab^2-b^3 for factoring x^3-9x^2+27x-27)

we have: a=1 and b=3, and x^3-12x^2+27x-27=(x-3)^3, so g(x)=(x-3)(x-3)^3=(x-3)^4

Thus, f(x)=xg(x)=x(x-3)^4.   That is: x^5-12x^4+54x^3-108x^2+81x=x(x-3)^4=0

We have: x=0 or x=3  

CK: Plug the values of x into Eq.1.   We have: f(0)=0 and f(3)=0   CKD.

Therefore, the answers are: x=0 or x=3