Question

Factor completely 3x4 – 6x3 + 12x2, Explain?

Answer 1

First, factor out any common factors, either numeric or in the variable.

Answer 2

What do you get if you do that?

Answer 3

3x^2(x^2-2x+4) ?

Answer 4

\[3x^4-6x3 +12x^2 = 3x^2(x^2-2x+4)\]

Now, can you factor the polynomial in the parentheses?

Answer 5

yes?

Answer 6

How does it factor?

Answer 7

3x2(x – 2)(x – 2) ?

Answer 8

hmm. what do you get if you multiply
\[(x-2)(x-2)\]

Answer 9

3x^2(x^2-4x+4) so it cant be facotr further?

Answer 10

In fact, this cannot be factored further, as \(x^2-2x+4\) is irreducible. So,
\[3x^4-6x^3+12x^2 = 3x^2(x^2-2x+4)\]is a complete factoring.

Answer 11

If we put \(x^2-2x+4=0\) through the quadratic formula, we see the solutions (which are the same as the factors, of course) are
\[x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(4)}}{2(1)} =\frac{4\pm\sqrt{4-16 }}{2} = 2\pm\sqrt{3} \]

Answer 12

so it could be factored as \[3x^2(x-(2+\sqrt{3}))(x-(2-\sqrt{3}))\]but typically that isn't done.

Answer 13

ah, crap, I forgot the \(i\)s! :-(
\[x = 2 \pm i\sqrt{3}\]
\[3x^2(x-(2+i\sqrt{3}))(x-(2-i\sqrt{3}))\]
Thought that was a little too easy to type :-)

Answer 14

thanks