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.
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.
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) = -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)
Method 1: Quadratic Equations in Standard Form
Video Courtesy of (site-affiliate) Kaplan GMAT 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.
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.
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.
(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.
2y2 – 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 + 6y – 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.
5y2 – 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
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.
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.
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.
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.
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.
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
Video Quiz
Quadratic Equations
Best viewed in landscape mode
4 questions with video explanations
100 seconds per question
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https://www.youtube.com/watch?v=z1hCqLclHQg
https://www.youtube.com/watch?v=jDiQunAKp28
https://www.youtube.com/watch?v=Bs7s7hGAWos
https://www.youtube.com/watch?v=GnZhjfINFjs
https://www.youtube.com/watch?v=zY1bPQqhWjc&list=PL6F58D00ADB3C0A85&index=7
Before attempting these problems, be sure to review this section on data sufficiency questions.
https://www.youtube.com/watch?v=hB_8gys2A4k
https://www.youtube.com/watch?v=rVYEtYEKp8Y