Algebra Tutorials!
Wednesday 18th of January
Solving Linear Equations
Systems of Linear Equations
Solving Linear Equations Graphically
Algebra Expressions
Evaluating Expressions and Solving Equations
Fraction rules
Factoring Quadratic Trinomials
Multiplying and Dividing Fractions
Dividing Decimals by Whole Numbers
Adding and Subtracting Radicals
Subtracting Fractions
Factoring Polynomials by Grouping
Slopes of Perpendicular Lines
Linear Equations
Roots - Radicals 1
Graph of a Line
Sum of the Roots of a Quadratic
Writing Linear Equations Using Slope and Point
Factoring Trinomials with Leading Coefficient 1
Writing Linear Equations Using Slope and Point
Simplifying Expressions with Negative Exponents
Solving Equations 3
Solving Quadratic Equations
Parent and Family Graphs
Collecting Like Terms
nth Roots
Power of a Quotient Property of Exponents
Adding and Subtracting Fractions
Solving Linear Systems of Equations by Elimination
The Quadratic Formula
Fractions and Mixed Numbers
Solving Rational Equations
Multiplying Special Binomials
Rounding Numbers
Factoring by Grouping
Polar Form of a Complex Number
Solving Quadratic Equations
Simplifying Complex Fractions
Common Logs
Operations on Signed Numbers
Multiplying Fractions in General
Dividing Polynomials
Higher Degrees and Variable Exponents
Solving Quadratic Inequalities with a Sign Graph
Writing a Rational Expression in Lowest Terms
Solving Quadratic Inequalities with a Sign Graph
Solving Linear Equations
The Square of a Binomial
Properties of Negative Exponents
Inverse Functions
Rotating an Ellipse
Multiplying Numbers
Linear Equations
Solving Equations with One Log Term
Combining Operations
The Ellipse
Straight Lines
Graphing Inequalities in Two Variables
Solving Trigonometric Equations
Adding and Subtracting Fractions
Simple Trinomials as Products of Binomials
Ratios and Proportions
Solving Equations
Multiplying and Dividing Fractions 2
Rational Numbers
Difference of Two Squares
Factoring Polynomials by Grouping
Solving Equations That Contain Rational Expressions
Solving Quadratic Equations
Dividing and Subtracting Rational Expressions
Square Roots and Real Numbers
Order of Operations
Solving Nonlinear Equations by Substitution
The Distance and Midpoint Formulas
Linear Equations
Graphing Using x- and y- Intercepts
Properties of Exponents
Solving Quadratic Equations
Solving One-Step Equations Using Algebra
Relatively Prime Numbers
Solving a Quadratic Inequality with Two Solutions
Operations on Radicals
Factoring a Difference of Two Squares
Straight Lines
Solving Quadratic Equations by Factoring
Graphing Logarithmic Functions
Simplifying Expressions Involving Variables
Adding Integers
Factoring Completely General Quadratic Trinomials
Using Patterns to Multiply Two Binomials
Adding and Subtracting Rational Expressions With Unlike Denominators
Rational Exponents
Horizontal and Vertical Lines
Try the Free Math Solver or Scroll down to Tutorials!












Please use this form if you would like
to have this math solver on your website,
free of charge.

Simple Trinomials as Products of Binomials

The Most General Trinomial

Both methods work only for trinomials in which the coefficient of the x 2 term is 1. More general trinomials can sometimes also be factored. Note that

(ax + b)(cd + d) = ac · x 2 + (ad + bc)x + bd

So, a general trinomial (one in which the coefficient of x 2 is a whole number different from 1) can be factored into the form of a product of two binomials, (ax + b)(cx + d), if we can find four whole numbers, a, b, c, and d, satisfying

ac = the coefficient of x 2

ad + bc = the coefficient of x


bd = the constant term

Analyzing these conditions can require quite a lengthy process, depending on the specific values of the various coefficients, because of the large number of possibilities that must be examined, and the complexity of the second condition.

The easiest approach is to make a table of sets of whole number values of a, b, c, and d, which satisfy the first and third conditions above. These are the most restrictive conditions, and so will narrow down the possibilities to be examined as much as possible. For each quadruple of values in the table, the value of ad + bc would then be calculated to determine which quadruple, if any of the possibilities, satisfies the second condition. If such a set of values is found, the factorization follows according to the pattern shown in the box above. If no such set of values is found, the trinomial cannot be factored in this way.

We illustrate this method with a simple example.

Example 1:

Factor 6x 2 + 13x – 5 as much as possible.


The three terms here contain no common monomial factors, so the only possibility is to try to factor this expression into the product of two binomials. Since the coefficient of x 2 is not equal to 1, we must seek a factorization of the form

(ax + b)(cx + d).

The numbers in this product must satisfy

ac = 6 (coefficient of x 2 )

bd = -5 (constant term)

ad + bc = 13 (coefficient of x).

Since ac = 6 is a positive value, we know that a and c must have the same signs. The positivevalued possibilities for a and c are

a = 1 and c = 6

a = 6 and c = 1

a = 2 and c = 3

a = 3 and c = 2

We don’t need to consider negative values of a and c as long as we include both positive and negative possibilities for b and d. The list of four possible pairs of values for a and c above might appear at first to duplicate each actual possibility. However, since a occurs in the factor with b, and c occurs in the factor with d, the situation a = 1 and c = 6 is actually different than the situation with a = 6 and c = 1.

The condition bd = -5 restricts possible values of b and d to the combinations

b = 1 and d = -5

b = -1 and d = 5

b = 5 and d = -1


b = -5 and d = 1.

Note that our list includes the positives and negatives of all values of b which divide evenly into -5. This is necessary to ensure that our analysis includes all possible combinations of whole numbers that might work here.

The four possible pairs of values for a and c can now be matched with each of the four possible pairs of values for b and d, giving a total of 16 possibilities to be check. (You can see that when the coefficient of x 2 and the constant term of the trinomial have a lot of whole number factors, the combined number of possibilities to be examined at this stage can become very large, so that this factorization can become a very tedious job – and often the end result is the conclusion that the trinomial cannot be factored!)

Anyway, setting the analysis up in a table gives:

Thus, it appears that

6x 2 + 13x – 5 = (2x + 5)(3x – 1)


(2x + 5)(3x – 1) = (2x + 5)(3x) + (2x + 5)(-1)

= (3x)(2x + 5) +(-1)(2x + 5)

= (3x)(2x) + (3x)(5) + (-1)(2x) + (-1)(5)

= 6x 2 + 15x – 2x 5

= 6x 2 + 13x – 5

as required. Our proposed factorization of the original trinomial has been verified.

Two, no three, final comments with regard to the method demonstrated with the example above:

(i) It is logically ok to sort of randomly test various possibilities in hope of finding an acceptable combination of values that might work without having to test all of the possibilities in the table. It is important to set the table up systematically however. If you just try random combinations of values of a, b, c, and d, without some way of keeping track of which combinations you’ve tested, you’ll end up going in circles, almost certainly. Of course, the only way to conclude that a trinomial of this type cannot be factored is to make sure you test all possible combinations of values of these four constants. (Yes, if you know how to solve a quadratic equation, you can achieve this factorization or demonstrate that no factorization is possible with quite a lot less work.)

(ii) You can see from the table of test cases above that we still have duplication. Try to figure out how to avoid at least some of the duplication that occurred above. (Hint: the product (ax + b)(cx + d) is identical to the product (cx + d)(ax + b).)

(iii) This is really a very tedious type of problem to solve even under the best of circumstances. There are more powerful and efficient methods for finding the factors of a trinomial which get around the need to enumerate large sets of possible factorizations, but these methods are also more complicated and beyond the scope of this lesson.

Copyrights © 2005-2017