How to factor $16x^2-y^2+8y-16$ so I get $(4x+y-4)(4x-y+4)$?

Question

I tried completing the square, making it equal to zero, and trying to think of it as a circle, but I still can't prove that these two are equivalent. I tried to factor this in Wolfram Alpha, and it spits out that answer. But I don't know how else to get to it.

Answer 1

We have that

$$16x^2-y^2+8y-16 =16x^2-(y^2-8y+16)=(4x)^2-(y-4)^2$$

then recall that

$$A^2-B^2=(A+B)(A-B)$$

Answer 2

$$16x^2-y^2+8y-16=16x^2-(y-4)^2=(4x-(y-4))(4x+(y-4))$$

At the last step I used $a^2-b^2=(a-b)(a+b)$, in which $a= 4x$ and $b=y-4$.

Answer 3

I f you want to be systematic, add in a variable $z$ to make this homogeneous, $$ 16 x^2 - y^2 - 16 z^2 + 8 y z $$ with result (below)
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z = 16 x^2 - (y-4z)^2 = (4x+y-4z)(4x-y+4z) $$

Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$

==============================================

$$ E_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 4 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & - 1 & 4 \\ 0 & 4 & - 16 \\ \end{array} \right) $$