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

## Definitions

#### Monomial

a number, variable, or a product of numbers and variables with whole number exponents.
In 2x4 – 5x2 + 3, each of 2x4, -5x2, and 3 is a monomial.

#### Polynomial

a monomial or sum of monomials.
2x4– 5x2 + 3 is a polynomial. Each monomial is called a term.

#### Degree of a polynomial

the exponent on the term that has the largest exponent. The degree of
2x4 – 5x2 + 3 is 4.

#### Binomial

a polynomial with two terms. 2x4 + 3 and y – 7 are binomials.

an equation of degree 2 that can be written in the form ax2 + bx + c = 0 where a, b, and c are constants, and
a ≠ 0.

#### Root

can refer to a solution of a polynomial equation, such as square root or cube root.

To add or subtract polynomials, add like terms. You can use a horizontal or vertical format.

##### Example

Simplify:

(a) (9x3 + 2x2 + x) + (7x2– 5x + 8)

### Solution

(a) (9x3 + 2x2 + x) + (7x2 – 5x + 8)
= 9x3 + 2x2 + 7x2 + x – 5x + 8
= 9x3 + 9x2 – 4x + 8

##### Example

Simplify:

(b) (y4 + 4y3 + 2y2y – 10)
– (7y4 + 4y3 + 5y + 12)

### Solution

(b) xxxy4 + 4y3 + 2y2y – 10
xxx–(7y4 + 4y3 xxxx+ 5y + 12)
xxxx-6y4 xxxxx+ 2y2 – 6y – 22

## Multiplying Polynomials

To multiply a monomial by a monomial, multiply the numerical coefficients and each variable. (To review rules of exponents, see the Exponents section in the previous chapter.)

##### Example

Simplify:   (2ps3)(-3p3s2)

### Solution

(2ps3)(-3p3s2)         …Multiply coefficients:   2 × -3 = -6

= -6 × p4× s5         …and each variable:     p × p3 = p4     s3 × s2 = s5

= -6p4s5

Multiplying polynomial expressions is based on the distributive property. To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial.

##### Example

Simplify: 4x(3x2 – 2xy + y2)

### Solution

#### Solution

4x(3x2 – 2xy + y2)
= 4x(3x2) – (4x)(2xy) + (4x)( y2)
= 12x3 – 8x2y + 4xy2

To multiply two polynomials, multiply each term of the first expression by each term of the second. Then combine like terms.

##### Example

Simplify: (x + 3)(3x2 – 4x + 2)

### Solution

(x + 3)(3x2 – 4x + 2)

x(3x2 – 4x + 2) + 3(3x2 – 4x + 2)
…Distribute x and 3 to
(3x2 – 4x + 2).

= (3x3 – 4x2 + 2x) + (9x2 – 12x + 6)      …Distribute to each term.

= 3x3 + 5x2 – 10x + 6                            …Combine like terms.

Finding area and volume often uses multiplication of polynomials.

##### Example

The dimensions of a box are
(y + 2) feet long, (y + 7) feet wide and (2y – 4) feet high. What is the area of the bottom of the box? What is the volume of the box?

### Solution

To find the area of the bottom of the container, multiply length by width, which are the first two binomials.

(y + 2)(y + 7)
y(y + 7) + 2(y + 7)
y2 + 9y + 14

To find the volume, multiply the area of the bottom by the height.

(2y – 4)(y2 + 9y + 14)

= (2y3 + 18y2 + 28y)
– (4y2 + 36y + 56)

= 2y3 + 14y– 8y – 56

## Multiplying Binomials

The most common polynomials on the GRE are binomials and quadratics. Multiplying two binomials gives a quadratic.

A quick way to multiply two binomials is to use FOIL (First, Outer, Inner, Last), then combine like terms.

First:

Multiply the first terms of each binomial.

Outer:

Multiply the outside terms: the first and last terms of the expression.

Inner:

Multiply the two terms closest to each other.

Last:

Multiply the last terms of each binomial.

##### Example

Multiply:   (x + 3)(x – 5)

### Solution

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

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

There are three common patterns in multiplying binomials. Memorizing these patterns will help in multiplying and factoring binomials and polynomials.

(a + b)2 = a2 + 2ab + b2
(x + 4)2 = x2 + 8x  + 16

(a – b)2 = a2 – 2ab + b2
(x – 4)2 = x2 – 8x  + 16

(a + b)(a – b) = a2 – b2
(x + 4)(x – 4) = x2 – 16   (called the difference of squares)

Multiply:

(a) (y – 9)2

### Solution

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

Multiply:

(b) (-7p + 2)2

### Solution

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

##### Example

Multiply:

(c) (3x + y)(3x – y)

### Solution

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

## Solving Quadratic Equations

Strategies for Quadratic Equations

Video Courtesy of Kaplan GRE prep.

quadratic equation is an equation of degree 2 that can be written in standard form
ax2 + bx + c = 0 where a ≠ 0.

Factoring is the simplest way to solve most quadratic equations. In a quadratic equation, one factor must equal zero.

You know that (x + 3)(x + 4)
x2 + (3 + 4)x + (3 × 4) = x2 + 7x + 12. Reverse this process to factor a quadratic equation x2 + bx + c = 0.

#### How to factor x2 + bx + c = 0:

(1)  If b and c are positive, then the two factors of c are positive and their sum is b.

Solve:  x2 + 11x + 18 = 0
Find two factors of 18 whose sum is 11.

### Solution

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.

(2)  If b is negative and c is positive, then the factors of c are negative and their sum is b.

Solve:  x2 – 6x + 8 = 0
Find two factors of 8 that are negative and whose sum is -6.

### Solution

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.

(3)  If c is negative, then the factors of c have different signs and their sum is b.

Solve:  x2 – 7x – 30 = 0
Find two factors of -30 whose sum is -7.

### Solution

Make a table of the factors of -30.

 1 -30 -1 30 2 -15 -2 15 3 -10 -3 10 5 -6 -5 6

One factor must be positive and one negative.
The sum must be -7.
3 + (-10) = -7, so x2 – 7x – 30
= (x + 3)(x – 10) = 0.
The solutions are
x = -3 and x = 10.

Make a table of the factors of -30.
One factor must be positive and one negative.
The sum must be -7.

3 + (-10) = -7, so x2 – 7x – 30 = (x + 3)(x – 10) = 0.
The solutions are x = -3 and x = 10.

 1 -30 -1 30 2 -15 -2 15 3 -10 -3 10 5 -6 -5 6
##### Example

Solve:  x2 – 14x + 49 = 0

### Solution

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.

## Quadratic Equations in Standard Form

Video Courtesy of Kaplan GRE prep.

x2 + 8x + 16 = 0 is clearly a quadratic equation. But quadratic equations are rarely in standard form. To solve, you may have to rewrite the equations in the form ax2 + bx + c = 0.

Solve:

(a) y2 = 5y

### Solution

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

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

Solve:

(b) y2 = 6 – y

### Solution

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

(b) 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.

##### Example

Solve:  2x – 3 = (4/x) + x

### Solution

2x – 3 = (4/x) + x      …Subtract x from both sides.

x – 3 = (4/x)        …Multiply both sides by x to get rid of the fraction.

x(x – 3) = x(4/x)        …Simplify.

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

x– 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.

##### Example

Solve: (152)2 – (148)2

### Solution

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

Factoring is also a method for solving quadratic equations ax2 + bx + c = 0 when a ≠ 1.    The process is to find pairs of factors of a and of c that add to get b.

##### Example

Solve:

(a) 2y– 7y + 3 = 0

### Solution

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

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

 factors of a negative factors of c value of b when multiplied 1, 2 -1, -3 (1 × -1) + (2 × -3) = -1 + -6 = -7 1, 2 -3, -1 (1 × -3) + (-1× 2) = -3 + -2 = -5

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 = 1/2.
y – 3 = 0, so y = 3.

The solutions are y = 1/2 and y = 3.

##### Example

Solve:

(b) 3y2 + 14y – 5 = 0

### Solution

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

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

 factors of a factors of c value of b when multiplied 1, 3 1, -5 (1 × 1) + (3 × -5) = 1 + -15 = -14 1, 3 -1, 5 (1 × -1) + (3 × 5) = -1 + 15 = 14

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 = 1/3.            y + 5 = 0, so y = -5.

The solutions are y = 1/3 and y = -5.

##### Example

Solve:

(c) -5y2 + 6– 1 = 0

### Solution

(c) -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

 factors of a factors of c value of b when multiplied 1, 5 1, 1 (1 × 1) + (1 × 5) = 1 + 5 = 6 1, 5 -1, -1 (1 × -1) + (5 × -1) = -1 + -5 = -6

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

5y– 6y + 1 = (5y – 1)(y – 1) = 0    The solutions are y = 1/5 and y = 1.

Solve:

(a) 4x2 – 81 = 0

### Solution

(a) 4x2 – 81 = 0 is the difference of squares, (2x)2 and 92.
4x2 – 81 = (2x – 9)(2x + 9) = 0, so
x = 9/2 and x = -9/2.

##### Example

Solve:

(b) 6x2 + 18x – 24 = 0

### Solution

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

(b) 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.

##### Example

Solve:

(c) x3 + x2 – 2x = 0

### Solution

(c) xx2 – 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.

##### Example

Factor:

xw + yw + zx + zy

### Solution

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)

##### Example

The sides of a rectangle are (2x – 9) feet long and (x – 1) feet wide.  The area of the rectangle is 72 square feet.  Find the value of x.

### Solution

(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 = -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.

## Solving Using the Quadratic Formula

Factoring is the best way to solve almost all equations on the GRE. Another method is to use the quadratic formula. It will solve a quadratic equation, but takes much more time and calculation. The solutions often contain square roots, so values will only be estimates.

The solutions of the quadratic equation ax2 + bx + c = 0, a ≠ 0, are:

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

Solve.

3x2 – 5x + 1 = 0

### Solution

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.

(5 ± \sqrt{13})/6 = (5 ± 3.5)/6.

So x ≈ (5 + 3.5)/6 = 8.5/6 ≈ 1.42 and
x ≈ (5 – 3.5)/6 = 1.5/6 = 0.25.

## Solving Radical Equations

radical equation is an equation with terms that are the square root of a variable. To solve a radical equation, isolate the radical on one side, then square both sides.

##### Example

Solve.

(a) 2\sqrt{\textit{x}} – 8 = 0

### Solution

(a) 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

##### Example

Solve.

(b) 4\sqrt{(\textit{x} \,-\, 8)} + 7 = 31

### Solution

(b) 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

##### Example

Solve.

(c) \sqrt{(21 \,-\, \textit{x})} \,-\, \sqrt{(\textit{x} \,-\, 1)}  = 0

### Solution

(c) 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

#### Be Careful!

In solving radical equations you are squaring both sides,
using the property: If a = b, then a2 = b2.
There is one answer.

Find the value of x2 when x = -10.
x2 = (-10)2 = 100

BUT taking a square root has two answers,
a positive and a negative square root:
If a2 = b2, then a = ± b.

Solve:  x2 = 49
x = ± 7
There are two solutions: x = 7 and
x = -7.

Solve.

(x – 3)2 = 36

### Solution

(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

#### Exponent Expressions

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