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.
Quadratic equation
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.
Adding Polynomials
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 + 2y2 – y – 10)
– (7y4 + 4y3 + 5y + 12)
Solution
(b) xxxy4 + 4y3 + 2y2 – y – 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 + 14y2 – 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) = -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
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)
Example
Multiply:
(a) (y – 9)2
Solution
(a) (y – 9)2
= y2 – 2y(9) + 81 …Use (a – b)2
= a2– 2ab + b2.
= y2 – 18y + 81
Example
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.
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.
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.
Example
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.
Example
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.
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.
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) 2y2 – 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 + 6y – 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.
5y2 – 6y + 1 = (5y – 1)(y – 1) = 0 The solutions are y = 1/5 and y = 1.
Example
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) 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.
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}}
Example
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
A 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.
Example
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
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Exponent Expressions
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