Free GRE Course > GRE Math Basics > Algebra > Complex Expressions

Dividing by Monomials

You have already found the products of polynomials.

[[https://www.youtube.com/watch?v=i7w_aRlVagU]] a few secons in after intro

3y(y2 + 2y + 7) = 3y3 + 6y2 + 21y

(x + 3)(x + 2) = x2 + 5x + 6

You can use the same principles to divide polynomials.

dividing by a monomial

\dfrac{3\textit{y}^{\displaystyle{3}} + 6\textit{y}^{\displaystyle{2}} + 21\textit{y}}{3\textit{y}} = y2 + 2y + 7

dividing by a binomial

\dfrac{\textit{x}^{\displaystyle{2}} + 5\textit{x} + 6}{\textit{x} + 3} = x + 2

Two methods for dividing a polynomial by a monomial are factoring and dividing each term.

Example

Simplify:

(a) \dfrac{(6\textit{x}^{\displaystyle{3}} -\,8\textit{x}^{\displaystyle{2}} + 12\textit{x})}{2\textit{x}}

Solution

(a) 6x3 – 8x2 + 12x
Look for common factors.
6, 8, and 12 have the common factor 2. The common factor of x3x2 and x is x.
6x3 – 8x2 + 12x = 2x(3x2 – 4x + 6)
(6x3 – 8x2 + 12x) ÷ 2x = 3x2 – 4x + 6

Example

Simplify:

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

Solution

(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

Dividing By Binomials

Two methods for dividing a polynomial by a binomial are factoring and long division.

Example

Simplify:

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

Solution

(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

Example

Simplify:

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

Solution

(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}

Example

Simplify:

(c) \dfrac{(6\textit{y}^{\displaystyle{2}} + 7\textit{y} -\, 3)}{(3\textit{y} -\, 1)}

Solution

(c) Use the format of long division:

—————2y + 3
3y – 1\overline{ \big)\,6y^{\displaystyle{2}} + 7y \,-\, 3}
xxxxxxxxx6y2 – 2y
xxxxxxxxxxx9y – 3
xxxxxxxxxxx9y – 3
xxxxxxxxxxxxxxx0

6y2 ÷ 3y = 2y
Multiply
3y – 1
by 2y and subtract.
7y – (-2y)
= 9y
9y ÷ 3y = 3
Multiply
3y – 1
by 3 and subtract.

(6y2 + 7y – 3) ÷ (3y – 1) = 2y + 3

Complex Fractions

complex fraction is a fraction that has a fraction in the numerator or denominator. In other words, it is a fraction divided by a fraction. Complex fractions can contain variable expressions. To simplify, multiply by the reciprocal, then factor and cancel any common factors.

Example

Simplify:

(a) \dfrac{\dfrac{\textit{x}^{\displaystyle{2}}}{9}}{\dfrac{3\textit{x}}{4}}

Solution

(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}

Example

Simplify:

(b) \dfrac{\dfrac{9\textit{y}^{\displaystyle{2}}}{7}}{12\textit{y}}

Solution

(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.

Example

Simplify:

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

Solution

(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.

Adding Polynomial Fractions

When adding (or subtracting) algebraic fractions, follow the same process as adding number fractions. The first step is to write equivalent fractions that have the same (common) denominator.

  1. Find the least common denominator (LCD) of the fractions.
  2. Write equivalent fractions using the LCD.
  3. Add (or subtract) the numerators.
  4. Simplify and reduce the resulting fraction.

A quick method for finding the LCD is to multiply the denominators. But multiplying the denominators often gives a rather large expression. Instead of multiplying, factor the denominators. Use the least common multiple. The LCM will have one of each shared factor and all of the factors that are not shared. (To review LCD and LCM, see the sections on Divisibility and Fractions in the previous chapter.)

Example

Find the least common multiple:

(a) 14s and 6s2

Solution

(a) Factor.

142 × 7× s
6s2 = 2 × 3 × s × s

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

Example

Find the least common multiple:

(b) y2– 25 and y2 + 6y + 5

Solution

(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).

Example

Solve: \dfrac{1}{2} + \dfrac{1}{2\textit{x}} + \dfrac{3}{\textit{x}^{\displaystyle{2}}} = 1

Solution

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 

Example

Solve: \dfrac{5\textit{y}}{\textit{y} \,-\, 2} = \dfrac{15}{\textit{y}} + \dfrac{10}{\textit{y} \,-\, 2}

Solution

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.

How Many Solutions?

The number of solutions for polynomial equations can vary. Some equations have infinite solutions or no solutions. Therefore, you need to be careful when performing operations on an equation that you don’t lose a possible solution or get an extraneous solution, which is a solution that does not satisfy the original equation. The value y = 2 in the Example above is an extraneous solution.

You might lose a solution if you divide both sides by the variable, and you might get an extraneous solution if you square both sides of an equation.

Example

Solve: \sqrt{5\textit{x}} + 5 = 0

Solution

\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.

Example

Solve:

(a) 4x3 = x

Solution

(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.

Example

Solve:

(b) 16 – y2 = 10(4 + y)

Solution

(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.

Example

Solve:

\sqrt{2\textit{x}} = x – 4

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 = 2then \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.

Example

Solve:

If \dfrac{\textit{x}}{\textit{y}} = 16 and \dfrac{\textit{x}}{\textit{y}^{\displaystyle{2}}} = 8, what is xy?

Solution

\dfrac{\textit{x}}{\textit{y}} = 16    …Multiply both sides by y.
x = 16y

\dfrac{\textit{x}}{\textit{y}^{\displaystyle{2}}} = 8    …Multiply both sides by y2.
x = 8y2

16y = 8y2   …Set the two values of x equal.

2y = y2
y2 – 2y = 0      y(y – 2) = 0
y = 2  and y = 0

You are solving for xy.   Substitute the values of y in either equation to find x.

\dfrac{\textit{x}}{\textit{y}} = 16 and y = 2
\dfrac{\textit{x}}{2} = 16            x = 32

\dfrac{\textit{x}}{\textit{y}} = 16 and y = 0
\dfrac{\textit{x}}{0} = 16    …Division by zero is undefined, so y = 0 is an extraneous solution.

The only solution is
xy = 32 × 2 = 64.

\dfrac{\textit{x}}{\textit{y}} = 16        Multiply both sides by y.        x = 16y

\dfrac{\textit{x}}{\textit{y}^{\displaystyle{2}}} = 8        Multiply both sides by y2.        x = 8y2

16y = 8y2      Set the two values of x equal.

2y = y2            y2 – 2y = 0      y(y – 2) = 0            y = 2  and y = 0

You are solving for xy.   Substitute the values of y in either equation to find x.

\dfrac{\textit{x}}{\textit{y}} = 16 and y = 2            \dfrac{\textit{x}}{2} = 16            x = 32

\dfrac{\textit{x}}{\textit{y}} = 16 and y = 0            \dfrac{\textit{x}}{0} = 16            Division by zero is undefined, so y = 0 is an extraneous solution.

The only solution is xy = 32 × 2 = 64.

Polynomials and Radicals

You can apply the properties of multiplying and dividing polynomials to simplify equations with radicals. Remember that the simplest form of an expression does not have radicals in the denominator. (To review rules of radicals, see the section on Roots in the previous chapter.)

Example

Simplify:

(a) \dfrac{6}{2\sqrt{3}}

Solution

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}

Example

Simplify:

(b) \sqrt{\dfrac{\textit{x}^{\displaystyle{2}}}{2}}

Solution

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}

Example

Simplify:

\big(\sqrt{\textit{x}} \,-\, \sqrt{\textit{y}}\big) \big(\sqrt{\textit{x}} + \sqrt{\textit{y}}\big)

Solution

\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

Variables as Exponents

Apply the properties of polynomials and exponents when exponents are variables. Rewrite terms to have the same base. Write an equation from the exponents. (To review, see the section on Exponents in the previous chapter.)

Example

Solve:

(a) 52y × 5– 1 = 25

Solution

(a) 52y × 5– 1 = 25       …Simplify each side of the equation.

52y × 5– 1 = 52y + y – 1 = 53y – 1       …On the left side of the equation, use am × an  = am + n.

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

53y – 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

Example

Solve:

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

Solution

(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 (am)n = amn.

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

21 – 2x = 25        …Substitute the simplified expressions in the original equation.

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

x = -2

Example

Factor: xa + xa + 1

Solution

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 xa + xa + 1 becomes x2 + x3. The common factor is x2, so x2 + x3 = x2(1 + x).

xa + xa + 1 = xa(1 + x)       
The common factor is xa.

Before attempting this problem, be sure to review this section on Quantitative Comparison questions.

https://www.youtube.com/watch?v=O1fMjpPFeks&list=PLD0D070C218D8F5A3&index=5

Find local GRE classes & schedules using our database of over 150 cities.