[Math] Factor $x^4 – 11x^2y^2 + y^4$

algebra-precalculusfactoring

This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.

The question is:

Factor $x^4 – 11x^2y^2 + y^4$

The answer is:

$(x^2 – 3xy -y^2)(x^2 + 3xy – y^2)$

My question is:

How did the textbook get this?

I tried the following methods (examples of my working below):

  1. U-Substitution.
  2. Guess and Check.
  3. Reversing the question (multiplying the answer out).

Here is my working for each case so far.

(1) U-Substitution.

I tried a simpler case via u-substitution.

Let $u = x^2$ and $v = y^2$. Then $x^4 – 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u – v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.

(2) Guess and Check.

To obtain the first term in $x^4 – 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:

$(x^2 \pm y^2)(x^2 \pm y^2)$.

But I'm at a loss as to how to get the final solution's middle term from guessing and checking.

(3) Reversing the question

Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.

The original answer:

$(x^2 – 3xy -y^2)(x^2 + 3xy – y^2) = [(x^2 – 3xy) – y^2][(x^2 + 3xy) – y^2]$

$= (x^2 – 3xy)(x^2 + 3xy) + (x^2 – 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$

$= [(x^2)^2 – (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 – 3y^3x] + [y^4]$

$= x^4 – 9x^2y^2 – y^2x^2 + 3y^3x – y^2x^2 – 3y^3x + y^4$

The $3y^3x$ terms cancel out, and we are left with:

$x^4 – 9x^2y^2 – 2x^2y^2 + y^4 = x^4 – 11x^2y^2 + y^4$, which is the original question.

The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x – 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x – 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).

Best Answer

\begin{align*} x^4 - 11x^2y^2 + y^4 &= x^4 - 2x^2y^2 + y^4-9x^2y^2 \\ &= (x^2-y^2)^2-(3xy)^2 \\ &= (x^2-y^2-3xy)(x^2-y^2+3xy). \end{align*}

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