Free Algebra
Tutorials!
 
Home
Point
Arithmetic Operations with Numerical Fractions
Multiplying a Polynomial by a Monomial
Solving Linear Equation
Solving Linear Equations
Solving Inequalities
Solving Compound Inequalities
Solving Systems of Equations Using Substitution
Simplifying Fractions 3
Factoring quadratics
Special Products
Writing Fractions as Percents
Using Patterns to Multiply Two Binomials
Adding and Subtracting Fractions
Solving Linear Inequalities
Adding Fractions
Solving Systems of Equations -
Exponential Functions
Integer Exponents
Example 6
Dividing Monomials
Multiplication can Increase or Decrease a Number
Graphing Horizontal Lines
Simplification of Expressions Containing only Monomials
Decimal Numbers
Negative Numbers
Factoring
Subtracting Polynomials
Adding and Subtracting Fractions
Powers of i
Multiplying and Dividing Fractions
Simplifying Complex Fractions
Finding the Coordinates of a Point
Fractions and Decimals
Rational Expressions
Solving Equations by Factoring
Slope of a Line
Percent Introduced
Reducing Rational Expressions to Lowest Terms
The Hyperbola
Standard Form for the Equation of a Line
Multiplication by 75
Solving Quadratic Equations Using the Quadratic Formula
Raising a Product to a Power
Solving Equations with Log Terms on Each Side
Monomial Factors
Solving Inequalities with Fractions and Parentheses
Division Property of Square and Cube Roots
Multiplying Two Numbers Close to but less than 100
Solving Absolute Value Inequalities
Equations of Circles
Percents and Decimals
Integral Exponents
Linear Equations - Positive and Negative Slopes
Multiplying Radicals
Factoring Special Quadratic Polynomials
Simplifying Rational Expressions
Adding and Subtracting Unlike Fractions
Graphuing Linear Inequalities
Linear Functions
Solving Quadratic Equations by Using the Quadratic Formula
Adding and Subtracting Polynomials
Adding and Subtracting Functions
Basic Algebraic Operations and Simplification
Simplifying Complex Fractions
Axis of Symmetry and Vertices
Factoring Polynomials with Four Terms
Evaluation of Simple Formulas
Graphing Systems of Equations
Scientific Notation
Lines and Equations
Horizontal and Vertical Lines
Solving Equations by Factoring
Solving Systems of Linear Inequalities
Adding and Subtracting Rational Expressions with Different Denominators
Adding and Subtracting Fractions
Solving Linear Equations
Simple Trinomials as Products of Binomials
Solving Nonlinear Equations by Factoring
Solving System of Equations
Exponential Functions
Computing the Area of Circles
The Standard Form of a Quadratic Equation
The Discriminant
Dividing Monomials Using the Quotient Rule
Squaring a Difference
Changing the Sign of an Exponent
Adding Fractions
Powers of Radical Expressions
Steps for Solving Linear Equations
Quadratic Expressions Complete Squares
Fractions 1
Properties of Negative Exponents
Factoring Perfect Square Trinomials
Algebra
Solving Quadratic Equations Using the Square Root Property
Dividing Rational Expressions
Quadratic Equations with Imaginary Solutions
Factoring Trinomials Using Patterns
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

Examples with solutions

Example 1:

Factor x 2 + 10x + 25 as much as possible.

solution:

The constant term here is a perfect square: 25 = 5 2 . Its square root is ± 5. But 2 · x · 5 = 10x, which happens to be the coefficient of x. So, we can write

x 2 + 10x + 25 = (x + 5) 2

You can check that this is correct by multiplying the right-hand side to remove the brackets, to confirm that the result obtained is the expression on the left-hand side.

Example 2:

Factor x 2 - 10x + 25 as much as possible.

solution:

This expression is very similar to the one in Example 1. The constant term is a perfect square, with square root of ± 5. In this case, the coefficient of the x term is -10 = 2 x (-5), so the trinomial has the form in the box above, but with a = -5. Thus, we conclude that

x 2 - 10x + 25 = (x – 5) 2

This is easily verified:

(x – 5) 2 = (x – 5)(x – 5)

= x(x – 5) + (–5)(x – 5)

= x 2 - 5x -5x + (–5) 2

= x 2 - 10x + 25

as required.

Example 3:

Factor x 2 - 10x - 25 as much as possible.

solution:

At first glance, this trinomial seems to have the same form as the trinomials in both Examples 1 and 2. However, matching this trinomial with the pattern shown in the box just above is not correct. Notice that in the template formula above, all terms are connected by ‘+’ signs. Thus, to match the trinomial in this example with that template, we need to write

x 2 -10x - 25 = x 2 -10x + (- 25)

Now we see that the constant term must be considered to be -25, not 25, and so it is not a perfect square. Thus, this trinomial cannot possibly match the pattern to potentially be equivalent to the square of a binomial.

You could still try to factor this trinomial using the more general method, to see if it can be factored into a product of the form (x + a)(x + b). The familiar table of possible values of a and b is:

a b a + b
1 25 24
5 5 0
1 25 24
5 5 0

Since none of the pairs of values which multiply to give -25 also sum to give -10, we conclude that no factorization of any form of this trinomial is possible.

Example 4:

Factor x 2 -10x + 25 as much as possible. Use the general method for factoring trinomials.

solution:

This is the same trinomial as considered in Example 2 above. In this case, we will not make use of the special form of the coefficients to recognize that it is equivalent to the square of a binomial. Instead, we’ll confirm that the general method for factoring simple trinomials will automatically generate that result.

So, if this trinomial is factorable, it will be into the form (x + a)(x + b), where

a + b = -10

and

ab = 25

The possible pairs of whole numbers which multiply to 25 are listed in the table

a b a + b
1 25 26
5 5 10
1 25 26
5 5 10

The fourth row of this table gives a sum of -10. Thus using a = -5 and b = -5 (that is, a = b = -5), we get

x 2 - 10x + 25 = (x + (-5)) (x + (-5)) = (x of the above – 5)(x – 5) = (x – 5) 2

as before.

All Right Reserved. Copyright 2005-2024