Binomials squared can be expanded using two main methods. We can write the binomial twice and perform the multiplication using the distributive property or we can use a standard formula and replace the values.
Here, we will look at a summary of these two methods for solving binomials squared. In addition, we will look at several examples with answers to understand how to apply these methods and fully master the binomials squared expansion.
Summary of binomials squared
A binomial squared is an expression that has the general form $latex {{(ax+b)}^2}$. This expression could contain other variables apart from x. For example, the expression $latex {{(5x+4y)}^2}$ is also a binomial squared.
There are two main methods that can be used to solve binomials squared:
First method
The first method consists in writing the binomial twice and eliminating the exponent. Then, we multiply the binomials using the distributive property or any other method. Finally, we combine like terms to simplify the resulting expression.
Second method
The second method consists of using a standard formula that tells us that the square of a binomial is equal to the sum of the square of the first term, twice the product of both terms, and the square of the last term.
Binomials squared – Examples with answers
The following examples use both of the methods detailed above to square the binomials. It is recommended that you try to solve the exercises yourself before looking at the solution.
EXAMPLE 1
Solve this binomial: $latex {{(2x+4)}^2}$.
Solution
Method 1: We rewrite the binomial as follows:
$latex{{(2x+4)}^2}$
⇒ $latex (2x+4)(2x+4)$
Now, we can multiply using the distributive property:
⇒ $latex 2x(2x+4)+4(2x+4)$
$latex =4{{x}^2}+8x+8x+16$
We combine like terms to simplify:
$latex =4{{x}^2}+16x+16$
Method 2: Using the standard formula, we have to find the square of the first term, twice the product of both terms, and the square of the last term:
⇒ $latex {{(2x)}^2}+2(2x)(4)+{{4}^2}$
Simplifying, we have:
⇒ $latex 4{{x}^2}+16x+16$
We got the same response with both methods, so both methods are valid.
EXAMPLE 2
Solve this binomial: $latex {{(3x-5)}^2}$.
Solution
Method 1: We eliminate the exponent and write the binomial twice:
$latex{{(3x-5)}^2}$
⇒ $latex (3x-5)(3x-5)$
With the distributive property, we can multiply and expand:
⇒ $latex 3x(3x-5)-5(3x-5)$
$latex =9{{x}^2}-15x-15x+25$
We simplify by combining like terms:
$latex =9{{x}^2}-30x+25$
Method 2: We find the square of the first term, twice the product of both terms and the square of the last term:
⇒ $latex {{(3x)}^2}+2(3x)(-5)+{{(-5)}^2}$
Simplifying, we have:
⇒ $latex 9{{x}^2}-30x+25$
EXAMPLE 3
Expand the binomial squared: $latex {{(-4x+10)}^2}$.
Solution
Method 1: We write the binomial twice and eliminate the exponent:
$latex{{(-4x+10)}^2}$
⇒ $latex (-4x+10)(-4x+10)$
We multiply using the distributive property:
⇒ $latex -4x(-4x+10)+10(-4x+10)$
$latex =16{{x}^2}-40x-40x+100$
We combine like terms to simplify:
$latex =16{{x}^2}-80x+100$
Method 2: To use this method we find the square of the first term, twice the product of both terms and the square of the last term:
⇒ $latex {{(-4x)}^2}+2(-4x)(10)+{{10}^2}$
Simplifying, we have:
⇒ $latex 16{{x}^2}-80x+100$
EXAMPLE 4
Solve this binomial squared: $latex {{(5x+2y)}^2}$.
Solution
Method 1: We eliminate the exponent and rewrite the binomial as follows:
$latex{{(5x+2y)}^2}$
⇒ $latex (5x+2y)(5x+2y)$
We expand and multiply to eliminate the parentheses:
⇒ $latex 5x(5x+2y)+2y(5x+2y)$
$latex =25{{x}^2}+10xy+10xy+4{{y}^2}$
Simplifying, we have:
$latex =25{{x}^2}+20xy+4{{y}^2}$
Method 2: To use the standard formula, we find the square of the first term, twice the product of both terms, and the square of the last term.:
⇒ $latex {{(5x)}^2}+2(5x)(2y)+{{(2y)}^2}$
Simplifying, we have:
⇒ $latex 25{{x}^2}+20xy+4{{y}^2}$
EXAMPLE 5
Solve this binomial squared: $latex {{(3{{x}^2}+4y)}^2}$.
Solution
Method 1: We rewrite the binomial as follows:
$latex{{(3{{x}^2}+4y)}^2}$
⇒ $latex (3{{x}^2}+4y)(3{{x}^2}+4y)$
Now, we multiply using the distributive property:
⇒ $latex 3{{x}^2}(3{{x}^2}+4y)+4y(3{{x}^2}+4y)$
$latex =9{{x}^4}+12{{x}^2}y+12{{x}^2}y+16{{y}^2}$
Simplifying, we have:
$latex =9{{x}^4}+24{{x}^2}y+16{{y}^2}$
Method 2: We find the square of the first term, twice the product of both terms, and the square of the last term to use the standard formula:
⇒ $latex {{(3{{x}^2})}^2}+2(3{{x}^2})(4y)+{{(4y)}^2}$
Simplifying, we have:
⇒ $latex 9{{x}^4}+24{{x}^2}y+16{{y}^2}$
EXAMPLE 6
Expand the binomial: $latex {{(5{{x}^2}-2{{y}^2})}^2}$.
Solution
Method 1: We eliminate the exponent and write the binomial twice:
$latex{{(5{{x}^2}-2{{y}^2})}^2}$
⇒ $latex (5{{x}^2}-2{{y}^2})(5{{x}^2}-2{{y}^2})$
We use the distributive property to multiply:
⇒ $latex 5{{x}^2}(5{{x}^2}-2{{y}^2})-2{{y}^2}(5{{x}^2}-2{{y}^2})$
$latex =25{{x}^4}-10{{x}^2}{{y}^2}-10{{x}^2}{{y}^2}+4{{y}^4}$
Simplifying, we have:
$latex =25{{x}^4}-20{{x}^2}{{y}^2}-4{{y}^4}$
Method 2: We find the square of the first term, twice the product of both terms, and the square of the last term to use the standard formula:
⇒ $latex {{(5{{x}^2})}^2}+2(5{{x}^2})(-2{{y}^2})+{{(-2{{y}^2})}^2}$
Simplifying, we have:
⇒ $latex 25{{x}^4}-20{{x}^2}{{y}^2}+4{{y}^4}$
Binomials squared – Practice problems
Practice what you have learned with the following problems. Expand the binomials to the square and choose an answer. If you need help, you can look at the solved exercises above.
See also
Interested in learning more about factoring and the quadratic formula? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
