对下列二次表达式进行配方。

1. Group the $x$ terms and complete the square by adding and subtracting the square of half the coefficient of $x$.

$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1$

2. Simplify the constant terms.

$= \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + 1$
$= \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$

1. Add and subtract the square of half the coefficient of $x$

$x^2 + 4x - 5 = x^2 + 4x + \left(\frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2 - 5$

2. Factorise the perfect square and simplify the constant terms

$= (x + 2)^2 - 2^2 - 5$
$= (x + 2)^2 - 4 - 5$
$= (x + 2)^2 - 9$

1. Factor out the coefficient of $x^2$ from the terms involving $x$.

$3x^2 + x - 1 = 3\left(x^2 + \frac{1}{3}x\right) - 1$

2. Complete the square inside the bracket by adding and subtracting the square of half the coefficient of $x$.

$= 3\left[ x^2 + \frac{1}{3}x + \left(\frac{1}{6}\right)^2 - \left(\frac{1}{6}\right)^2 \right] - 1$

$= 3\left[ \left(x + \frac{1}{6}\right)^2 - \frac{1}{36} \right] - 1$

3. Distribute the outer factor and simplify the constant terms.

$= 3\left(x + \frac{1}{6}\right)^2 - 3\left(\frac{1}{36}\right) - 1$

$= 3\left(x + \frac{1}{6}\right)^2 - \frac{1}{12} - 1$

$= 3\left(x + \frac{1}{6}\right)^2 - \frac{13}{12}$

1. Factor out the coefficient of $x^2$ from the $x^2$ and $x$ terms.

$-2x^2 - 4x + 3 \\ = -2(x^2 + 2x) + 3$

2. Complete the square for the expression inside the bracket.

$= -2\left[(x + 1)^2 - \left(\frac{2}{2}\right)^2\right] + 3 \\ = -2[(x + 1)^2 - 1] + 3$

3. Expand the outer bracket and simplify to obtain the final expression.

$= -2(x + 1)^2 + 2 + 3 \\ = -2(x + 1)^2 + 5$