First:              x × x = x²
Outer:            x × (-5) = -5x                            …Remember to use negative 5:  x – 5 = x + (-5)
Inner:            3 × x = 3x
Last:              3 × -5 = -15

Sum of F + O + I + L
= x² – 5x + 3x – 15       …Combine like terms -5x and +3x.
= x² – 2x  – 15

(y – 9)²
= y² – 2y(9) + 81     …Use (a – b)²
= a²– 2ab + b².
= y² – 18y + 81

(-7p + 2)²
= (-7p)² + 2(-7p)(2) + 2²   …Use (a + b)² 
= a² + 2ab + b².
= 49p² – 28p + 4

(3x + y)(3x – y)       = (3x)² – y² 
= 9x² – y²    …Use (a + b)(a – b)
= a² – b².

2x – 3 = (\dfrac{4}{\textit{x}}) + x      …Subtract x from both sides.

x – 3 = (\dfrac{4}{\textit{x}})        …Multiply both sides by x to get rid of the fraction.

x(x – 3) = x(\dfrac{4}{\textit{x}})        …Simplify.

x² – 3x = 4        …Subtract 4 from both sides.

x² – 3x – 4 = 0        …The factors of 4 are 1 and 4, and 2 and 2. There needs to be one positive factor, one negative factor and a sum of -3.

(x + 1)(x – 4) = 0        …The solutions are x = -1 and x = 4.

Though this equation is all numbers, don’t leap into doing the algebra. The calculations would take too long, so there must be a trick.

Notice it is a quadratic equation that is the difference of two squares.

Use a²– b² = (a + b)(a – b).

152² – 148² = (152 + 148)(152 – 148) = (300)(4) = 1,200

All of these equations need to be rewritten in standard form.

y² = 5y, so y² – 5y = 0.   …Factor out the y.
y(y – 5) = 0.
The solutions are y = 0 and y = 5.

All of these equations need to be rewritten in standard form.

y² = 6 – y, so y² + y – 6 = 0.
The factors of 6 are 1 and 6, and 2 and 3.
There needs to be one positive factor, one negative factor and a sum of 1.
(y + 3)(y – 2) = 0
The solutions are y = -3 and y = 2.

The factors of 18 are 1 and 18, 2 and 9, and 3 and 6.
The sum of 2 and 9 is 11, so 
x² + 11x + 18 = (x + 2)(x + 9) = 0.
The solutions are x = -2 and x = -9.

The negative factors of 8 are -1 and -8, and -2 and -4.
The sum of -2 and -4 is -6, so 
x² – 6x + 8 = (x – 2)(x – 4) = 0.
The solutions are x = 2 and x = 4.

Factor.  This equation is 
a² – 2ab + b² = (a – b)².

x² – 14x + 49 = (x – 7)²

Not all quadratic equations have two solutions. This equation has only one solution, x = 7.

2y² – 7y + 3 = 0             
a = 2   b = -7   c = 3

Since b is negative and c is positive, both factors of c are negative.

For the sum b to be -7, the factor pairs need to be 1 × -1 and 2 × -3.

2y² – 7y + 3 = (2y – 1)(y – 3) = 0

2y – 1 = 0, so y = \dfrac{1}{2}.       
y – 3 = 0, so y = 3.

The solutions are y = \dfrac{1}{2} and y = 3.

3y² + 14y – 5 = 0             
a = 3   b = 14   c = -5

Since c is negative, the factors of c have different signs.

For the sum b to be 14, the factor pairs need to be 1 × -1 and 3 × 5.

3y² + 14y – 5 = (3y – 1)(y + 5) = 0
3y – 1 = 0, so 3y = 1 and y = \dfrac{1}{3}.            y + 5 = 0, so y = -5.

The solutions are y = \dfrac{1}{3} and y = -5.

-5y² + 6y – 1 = 0

Since a is negative, first factor -1 from each term then divide both sides by -1.

-5y² + 6y – 1 = -(5y² – 6y + 1) = 0

For the sum b to be -6, the factor pairs need to be 1 × -1 and 5 × -1.

5y² – 6y + 1 = (5y – 1)(y – 1) = 0    The solutions are y = \dfrac{1}{5} and y = 1.

4x² – 81 = 0 is the difference of squares, (2x)² and 9².
4x² – 81 = (2x – 9)(2x + 9) = 0, so 
x = \dfrac{9}{2} and x = \dfrac{-9}{2}.

Notice the terms of the equation have the common factor 6.  Factor 6 from each term and divide both sides by 6.

6x² + 18x – 24 = 6(x² + 3x – 4) = 0

Solve x² + 3x – 4 = 0.

x² + 3x – 4 = (x + 4)(x – 1) = 0, so 
x = -4 and x = 1.

x³ + x² – 2x = 0   …Factor x from each term of the equation. Don’t divide both sides by x or you will lose the solution x = 0.

x(x² + x – 2) = 0     …Factor x² + x – 2.

x(x + 2)(x – 1) = 0      …The solutions are x = 0,  x = -2 and x = 1.

xw + yw + zx + zy        …Look for groups that share factors.

= (xw + yw) + (zx + zy)        …Each pair has the factor x + y.

= w(x + y) + z(x +  y)

= (w + z)(x +  y)

(2x – 9)(x – 1) = 72        …Write the quadratic equation.

2x²– 11x + 9 = 72        …Subtract 72 from both sides.

2x²– 11x – 63 = 0        …Factor.

(2x + 7)(x – 9) = 0

2x + 7 = 0    so x = \dfrac{-7}{2}.
x – 9 = 0      so x = 9.

The lengths of the sides of the rectangle cannot be negative numbers, so the only solution is
x = 9.

Looking at the factors of a = 3 and c = 1, no combination adds to b = -5 that is needed to factor the equation. So we must use the quadratic formula.

3x² – 5x + 1 = 0 , so a = 3, b = -5, and c = 1.

x = \dfrac{-\textit{b} \pm \sqrt{\textit{b}^{\displaystyle{2}} \,-\, 4\textit{ac}}}{2\textit{a}}

= \dfrac{-(-5) \pm \sqrt{(-5)^{\displaystyle{2}} \,-\, 4(3)(1)}}{2(3)}

= \dfrac{5 \pm \sqrt{25 \,-\, 12}}{6} = \dfrac{5 \pm \sqrt{13}}{6}

13 is between 9 and 16, so \sqrt{13} is between 3 and 4. Use the estimate \sqrt{13} ≈ 3.5.

\dfrac{(5 ± \sqrt{13})}{6} = \dfrac{(5 ± 3.5)}{6}.

So x ≈ \dfrac{(5 + 3.5)}{6} = \dfrac{8.5}{6} ≈ 1.42 and
x ≈ \dfrac{(5 \,-\, 3.5)}{6} = \dfrac{1.5}{6} = 0.25.

2\sqrt{\textit{x}} – 8 = 0      …Add 8 to both sides to get the radical alone on one side.
2\sqrt{\textit{x}} = 8     …Divide both sides by 2.
\sqrt{\textit{x}} = 4       …Square both sides.
(\sqrt{\textit{x}})² = 16      …Simplify.
x = 16

4\sqrt{(\textit{x} \,-\, 8)} + 7 = 31       …Subtract 7 from both sides to get the radical alone on one side.
4\sqrt{(\textit{x} \,-\, 8)}  = 24       …Divide both sides by 4.
\sqrt{(\textit{x} \,-\, 8)} = 6       …Square both sides.
x – 8  = 36       …Add 8 to both sides.
x = 44

For an equation with two radical expressions, put one radical expression on each side of the equation.
\sqrt{(21 \,-\, \textit{x})} \,-\, \sqrt{(\textit{x} \,-\, 1)} = 0       …Add \sqrt{(1 \,-\, \textit{x})} to both sides.
\sqrt{(21 \,-\, \textit{x})} = \sqrt{(\textit{x} \,-\, 1)}       …Square both sides.
21 – x = x – 1       …Add x and 1 to both sides.
22 = 2x       …Divide both sides by 2.
x = 11

(x – 3)² = 36        …Take the square root of both sides.

x – 3 = ± 6        …Add 3 to both sides.

x = 3 ± 6  →  x = 3 + 6 and x = 3 – 6.

The two solutions are x = 9 and 
x = -3