Free GMAT Course > GMAT Math Basics > 2. 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.

Multiplying Binomials

The most common polynomials on the GMAT 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.

Video Courtesy of (site-affiliate) Kaplan GMAT prep. $200 off Kaplan Tutoring 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. (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) (y – 9)2 Solution (y – 9)2 = y2 – 2y(9) + 81 …Use (ab)2 = a2– 2ab + b2. = y2 – 18y + 81 (-7p + 2)2 Solution (-7p + 2)2 = (-7p)2 + 2(-7p)(2) + 22 …Use (a + b)2 a2 + 2ab + b2. = 49p2 – 28p + 4 (3x + y)(3x – y) Solution (3x + y)(3x – y) = (3x)2 – y2 = 9x2 – y2 …Use (a + b)(a – b) a2 – b2. Method 1: Quadratic Equations in Standard Form Video Courtesy of (site-affiliate) Kaplan GMAT prep.$200 off Kaplan Tutoring

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.

2x – 3 = (\dfrac{4}{\textit{x}}) + x

Solution

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.

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.

(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

y2 = 5y

Solution

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.

y2 = 6 – y

Solution

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.

Method 2: Factoring

A 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.

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.

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.

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

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.

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.

2y– 7y + 3 = 0

Solution

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 = \dfrac{1}{2}.
y – 3 = 0, so y = 3.

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

3y2 + 14y – 5 = 0

Solution

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 = \dfrac{1}{3}.            y + 5 = 0, so y = -5.

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

-5y2 + 6– 1 = 0

Solution

-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 = \dfrac{1}{5} and y = 1.

4x2 – 81 = 0

Solution

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}.

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.

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

Solution

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.

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)

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 = \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.

Method 3: Solving Using the Quadratic Formula

Factoring is the best way to solve almost all equations on the GMAT. 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}}

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.

\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.

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.

2\sqrt{\textit{x}} – 8 = 0

Solution

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

Solution

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

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

Solution

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.

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.

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

(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