Completing squares with three variables.

Question

I want to complete the squares for this polynomial $2x^2+2y^2-z^2+2xy+3xz-4yz$

Is there any kind of easy and non-confusing way to solve it? I’ve done this up until now:

$$2x^2+2y^2-z^2+2xy+3xz-4yz$$ $$2x^2+2xy+3xz-4yz+2y^2-z^2$$ $$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$ $$2(x^2+x(y+\frac{3}{2}z))-4yz+2y^2-z^2$$ $$2(x^2+x(y+\frac{3}{2}z))+(\frac{1}{2}(y+\frac{3}{2}z))^2-(\frac{1}{2}(y+\frac{3}{2}z))^2-4yz+2y^2-z^2$$

Then things get kinda messy from here and I get totally lost from then on, could anyone help me out factoring this? And telling me if there is an eaay and non-confusing way to solve it?

Answer 1

Original is $$2x^2+2y^2-z^2-4yz+3zx+2xy$$

Fine until $$2(x^2+xy+\frac{3}{2}xz)-4yz+2y^2-z^2$$ at which point you want to stop, and calculate $$ \left(x + \frac{y}{2} + \frac{3z}{4} \right)^2 = x^2 + \frac{y^2}{4} + \frac{9 z^2}{16} + \frac{3yz}{4} + \frac{3zx}{2} + xy, $$ so $$ 2 \left(x + \frac{y}{2} + \frac{3z}{4} \right)^2 = 2x^2 + \frac{y^2}{2} + \frac{9 z^2}{8} + \frac{3yz}{2} + 3zx + 2xy \; \; .$$ $$2x^2+2y^2-z^2-4yz+3zx+2xy - 2 \left(x + \frac{y}{2} + \frac{3z}{4} \right)^2 = \frac{3y^2}{2} - \frac{17 z^2}{8} - \frac{11yz}{2}$$

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

Here is a way to do it with matrices, the first of which is the Hessian matrix of second partial derivatives of the function:

The final matrix equation, $Q^T DQ = H$ gives $$2x^2+2y^2-z^2-4yz+3zx+2xy = 2 \left(x + \frac{y}{2} + \frac{3z}{4} \right)^2 + \frac{3}{2} \left(y - \frac{11z}{6} \right)^2 - \frac{43}{6} z^2 \; \; . $$

$$ P^T H P = D $$ $$ Q^T D Q = H $$ $$ H = \left( \begin{array}{rrr} 4 & 2 & 3 \\ 2 & 4 & - 4 \\ 3 & - 4 & - 2 \\ \end{array} \right) $$

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

$$\left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 4 & 0 & 3 \\ 0 & 3 & - \frac{ 11 }{ 2 } \\ 3 & - \frac{ 11 }{ 2 } & - 2 \\ \end{array} \right) $$

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

$$\left( \begin{array}{rrr} 1 & 0 & - \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 3 }{ 4 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 3 & - \frac{ 11 }{ 2 } \\ 0 & - \frac{ 11 }{ 2 } & - \frac{ 17 }{ 4 } \\ \end{array} \right) $$

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

$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 11 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 5 }{ 3 } \\ 0 & 1 & \frac{ 11 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 11 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - \frac{ 43 }{ 3 } \\ \end{array} \right) $$

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

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - \frac{ 5 }{ 3 } & \frac{ 11 }{ 6 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 2 & 3 \\ 2 & 4 & - 4 \\ 3 & - 4 & - 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 5 }{ 3 } \\ 0 & 1 & \frac{ 11 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - \frac{ 43 }{ 3 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ \frac{ 3 }{ 4 } & - \frac{ 11 }{ 6 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - \frac{ 43 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 11 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 4 & 2 & 3 \\ 2 & 4 & - 4 \\ 3 & - 4 & - 2 \\ \end{array} \right) $$

Answer 2

If you mean writing polynomial as the sum of squares, here is one solution:

$p=2x^2+2y^2-z^2+2xy+3xz-4yz$

polynomial p can be written as following forms:

$p=(x+y)^2+(x+z)^2+(y-2z)^2-5z^2+xz$

$p=(x+y)^2 +(x+2z)^2+(y-2z)^2-5z^2 -xz$

Summing these relations and dividing the result by 2 we get:

$p= (x+y)^2 +\frac{1}{2}(x+z)^2+\frac{1}{2}(x+2z)^2 +(y-2z)^2 -(5^{1/2}z)^2$

Answer 3

We can do it simply using $(a \pm b )^2 = a^2 \pm 2ab + b^2$

$$2x^2 + 2y^2 -z^2 + 2xy +3xz - 4yz $$

$$ x^2 + y^2 + 2xy + x^2 +2(x)(\frac{3}{2}z)+(\frac{3}{2}z)^2 -(\frac{3}{2}z)^2+y^2 -2(y)(2z)+(2z)^2 -(2z)^2-z^2$$

$$(x+y)^2 +(x+\frac{3}{2}z)^2 +(y-2z)^2 - (\frac{\sqrt{29}}{2}z)^2 $$