(a) 6x³ – 8x² + 12x
Look for common factors.
6, 8, and 12 have the common factor 2. The common factor of x³, x² and x is x.
6x³ – 8x² + 12x = 2x(3x² – 4x + 6)
(6x³ – 8x² + 12x) ÷ 2x = 3x² – 4x + 6

(b) Divide each term.

\dfrac{25\textit{y}^{\displaystyle{4}} + 30\textit{y}^{\displaystyle{2}} \,-\, 20\textit{y}}{-5\textit{y}}

= \dfrac{25\textit{y}^{\displaystyle{4}}}{-5\textit{y}} + \dfrac{30\textit{y}^{\displaystyle{2}}}{-5\textit{y}} + \dfrac{-20\textit{y}}{-5\textit{y}}

= -5y³ – 6y + 4

(a) Factor.  \dfrac{\textit{x}^{\displaystyle{2}} \,-\, 3\textit{x} \,-\, 4}{\textit{x} + 1}

= \dfrac{(\textit{x} \,-\, 4)(\textit{x} + 1)}{(\textit{x} + 1)} = x – 4

(b) Factor. \dfrac{\textit{x} \,-\, 7}{2\textit{x}^{\displaystyle{2}} \,-\, 11\textit{x} \,-\, 21}

= \dfrac{\textit{x} \,-\, 7}{(\textit{x} \,-\, 7)(2\textit{x} + 3)} = \dfrac{1}{2\textit{x} + 3}

(c) Use the format of long division:

(6y² + 7y – 3) ÷ (3y – 1) = 2y + 3

(a) Use the reciprocal and multiply.

\dfrac{\dfrac{\textit{x}^{\displaystyle{2}}}{9}}{\dfrac{3\textit{x}}{4}} = \dfrac{\textit{x}^{\displaystyle{2}}}{9} \times \dfrac{4}{3\textit{x}} = \dfrac{4\textit{x}}{27}

(b) Use the reciprocal and multiply.

\dfrac{\dfrac{9\textit{y}^{\displaystyle{2}}}{7}}{12\textit{y}} = \dfrac{9\textit{y}^{\displaystyle{2}}}{7} \times \dfrac{1}{12\textit{y}}    …Division by 12y is multiplication by its reciprocal \dfrac{1}{12\textit{y}}.

\dfrac{3 \times 3 \times \textit{y} \times \textit{y}}{7} \times \dfrac{1}{3 \times 4 \times \textit{y}} = \dfrac{3\textit{y}}{28}    …Factor, then cancel common factors.

(c) Use the reciprocal and multiply.

\dfrac{\dfrac{\textit{x}^{\displaystyle{2}} + 7\textit{x}}{4}}{\dfrac{\textit{x}^{\displaystyle{2}} \,-\, 49}{\textit{x}}} 

= \dfrac{\textit{x}^{\displaystyle{2}} + 7\textit{x}}{4} \times \dfrac{\textit{x}}{\textit{x}^{\displaystyle{2}} \,-\, 49}    …Multiply by the reciprocal.

\dfrac{\textit{x}(\textit{x} + 7)}{4} \times \dfrac{\textit{x}}{(\textit{x} + 7)(\textit{x} \,-\, 7)}

= \dfrac{\textit{x}^{\displaystyle{2}}}{4(\textit{x} \,-\, 7)}    …Factor, then cancel common factors.

(a) Factor.

The LCM of 14s and 6s² is
2 × 7× s × 3 × s = 42s²

(b) Factor.

y² – 25 = (y + 5)(y – 5)                      
y² + 6y + 5 = (y + 1)(y + 5)

The LCM of y² – 25 and 
y² + 6y + 5 is
(y + 5)(y – 5)(y + 1).

The least common denominator (LCD) is the least common multiple, 2x².

\dfrac{1}{2} + \dfrac{1}{2\textit{x}} + \dfrac{3}{\textit{x}^{\displaystyle{2}}} = 1       …Multiply both sides by the LCD.

2x²\Bigg[\dfrac{1}{2} + \dfrac{1}{2\textit{x}} + \dfrac{3}{\textit{x}^{\displaystyle{2}}}\Bigg] = 2x²       …Use the distributive property.

\dfrac{2\textit{x}^{\displaystyle{2}}}{2} + \dfrac{6\textit{x}^{\displaystyle{2}}}{2\textit{x}} + \dfrac{10\textit{x}^{\displaystyle{2}}}{\textit{x}^{\displaystyle{2}}} = 2x²       …Cancel common factors in each fraction.

x² + 3x + 10 = 2x²       …Subtract 2x² from both sides and multiply by -1.

x² – 3x – 10 = 0       …Factor.

(x – 5)(x + 2) = 0     
so x = 5 and x = -2 

The least common denominator is y × (y – 2) = y(y – 2).

y(y – 2) \Bigg[\dfrac{5\textit{y}}{\textit{y} \,-\, 2}\Bigg]

= y(y – 2) \Bigg[\dfrac{15}{\textit{y}} + \dfrac{10}{\textit{y} \,-\, 2}\Bigg]       …Multiply both sides by the LCD.  Cancel common factors.

5y² = 15(y – 2) + 10y       …Use the distributive property.

5y² = 15y – 30 + 10y       …Set equations equal to zero.

5y² – 25y + 30 = 0       …Divide by 5.

y² – 5y + 6 = 0       …Factor.

(y – 2)(y – 3) = 0     so y = 2 and y = 3.

Check the answers by substituting in the original equation. If y = 2, the denominator y – 2 is equal to zero. Division by zero is undefined. So the only solution is y = 3.

\sqrt{5\textit{x}} + 5 = 0        …Subtract 5 from both sides.
\sqrt{5\textit{x}} = -5        …It looks like the next step is to square both sides, so x = 5
But on the GRE, a square root is never a negative number. There are no solutions.

(a) 4x³ = x     …It looks like the way to solve the equation is to divide both sides by x.
4x³ = x       4x² = 1       x² = 1/4       
x = 1/2 and x = -1/2

But by dividing both sides by x, you lose a value for x.
4x³ = x       4x³ – x = 0       
x(4x² – 1) = 0       x(2x + 1)(2x – 1) = 0

So x = 0 is another solution.  The solutions are x = 0, x = 1/2  and 
x = -1/2.

(b) 16 – y² = 10(4 + y)     …Factor.

(4 + y)(4 – y) = 10(4 + y)     …It looks like the next step is to divide both sides by 4 + y.

4 – y = 10        y = 4 – 10 = -6

But by dividing both sides by 4 + y, you lose a value for y.

16 – y² = 10(4 + y)       
16 – y² = 40 + 10y       
y² + 10y + 24 = 0

(y + 6)(y + 4) = 0       
y = -6 and y = -4       
So y = -4 is another solution.

\sqrt{2\textit{x}} = x – 4    …Square both sides.

2x = (x – 4)² = x² – 8x + 16       
x² – 10x + 16 = 0       
(x – 2)(x – 8) = 0

x = 2  and x = 8

Check the answers by substituting in the original equation.

If x = 2, then \sqrt{2\textit{x}} = x – 4 becomes \sqrt{4} = 2 – 4 = -2 .  So x = 2 is an extraneous solution.

The only solution is x = 8.

To simplify, multiply by a fraction equal to 1 with numerator and denominator equal to the radical.

(a) \dfrac{6}{2\sqrt{3}} = \dfrac{6}{2\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{2 \times 3} = \sqrt{3}

To simplify, multiply by a fraction equal to 1 with numerator and denominator equal to the radical.

(b) \sqrt{\dfrac{\textit{x}^{\displaystyle{2}}}{2}} = \dfrac{\sqrt{\textit{x}^{\displaystyle{2}}}}{\sqrt{2}}     …First apply the rule of exponents (\dfrac{\textit{a}}{\textit{b}})ⁿ = \dfrac{\textit{a}^{\displaystyle{\textit{n}}}}{\textit{b}^{\displaystyle{\textit{n}}}}
\\[3ex]\dfrac{\sqrt{\textit{x}^{\displaystyle{2}}}}{\sqrt{2}} = \dfrac{|\textit{x}|}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{|\textit{x}| \times \sqrt{2}}{2}

\big(\sqrt{\textit{x}} \,-\, \sqrt{\textit{y}}\big) \big(\sqrt{\textit{x}} + \sqrt{\textit{y}}\big)= x – y      …Use
(a + b)(a – b) = a² – b²

(a) 5^2y × 5^{y – 1} = 25       …Simplify each side of the equation.

5^2y × 5^{y – 1} = 5^{2y + y – 1} = 5^{3y – 1}       …On the left side of the equation, use a^m × aⁿ  = a^m + n.

25 = 5²        …On the right side of the equation, factor 25 so both bases are 5.

5^{3y – 1} = 5²        …Use the simplified expressions to write the original equation.

3y – 1 = 2        …Write just the exponents in an equation and solve.

y = 1

(b) \dfrac{2}{4^{\displaystyle{\textit{x}}}} = 32

\dfrac{2}{4^{\displaystyle{\textit{x}}}} = \dfrac{2}{(2^{\displaystyle{2}})^{\displaystyle{\textit{x}}}} = \dfrac{2}{2^{\displaystyle{2\textit{x}}}} = 2^{\displaystyle{1 \,-\, 2\textit{x}}}        …Use (a^m)ⁿ = a^mn.

32 = 2⁵        …On the right side of the equation, factor 32 so both bases are 2.

2^{1 – 2x} = 2⁵        …Substitute the simplified expressions in the original equation.

1 – 2x = 5        …Write just the exponents in an equation and solve.

x = -2

To simplify, find the common factor. If the common factor isn’t clear, use the Plugging In method to try some numbers.

Let a = 2. Then x^a + x^{a + 1} becomes x² + x³. The common factor is x², so x² + x³ = x²(1 + x).

x^a + x^{a + 1} = x^a(1 + x)       
The common factor is x^a.