In algebra, moving from binomials like $(a + b)^2$ to trinomials like $(a + b + c)^2$ can seem intimidating. However, once you understand the pattern, it becomes a simple “plug-and-play” process.
The standard formula for the square of a sum of three terms is:
Why does this work?
If we treat $(a + b)$ as a single unit, we can use the basic binomial expansion:
$[(a+b) + c]^2 = (a+b)^2 + c^2 + 2(a+b)c$
Expanding $(a+b)^2$ gives $a^2 + b^2 + 2ab$.
Distributing $2c$ gives $2ac + 2bc$.
Combining them all results in the final identity.
Expand $(2x + 3y + 4z)^2$
Using the formula where $a=2x$, $b=3y$, and $c=4z$:
Square the terms: $(2x)^2 + (3y)^2 + (4z)^2 = 4x^2 + 9y^2 + 16z^2$
Add the double products: $2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x) = 12xy + 24yz + 16zx$
Result: $4x^2 + 9y^2 + 16z^2 + 12xy + 24yz + 16zx$
Expand $(x + \frac{2}{x} + x^2)^2$
Even with fractions, the rule remains the same:
Squares: $x^2 + (\frac{2}{x})^2 + (x^2)^2 \rightarrow x^2 + \frac{4}{x^2} + x^4$
Products: $2(x)(\frac{2}{x}) + 2(\frac{2}{x})(x^2) + 2(x^2)(x)$
Simplified Result: $x^4 + 2x^3 + x^2 + 4x + \frac{4}{x^2} + 4$
What happens if one of the terms is negative, such as $(a + b – c)^2$?
You don’t need a brand-new formula! Simply treat $c$ as $(-c)$ and apply the standard identity.
Key Rule: Squaring a negative number results in a positive ($(-c)^2 = c^2$), but the product terms involving that single negative variable will become negative.
Identity for $(a + b – c)^2$:
Example: Expand $(2x + 3y – 5z)^2$
Squares: $(2x)^2 + (3y)^2 + (-5z)^2 = 4x^2 + 9y^2 + 25z^2$
Products: $2(2x)(3y) + 2(3y)(-5z) + 2(-5z)(2x) = 12xy – 30yz – 20zx$
Final Answer: $4x^2 + 9y^2 + 25z^2 + 12xy – 30yz – 20zx$
| Expression | Expansion |
| $(a+b+c)^2$ | $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ |
| $(a+b-c)^2$ | $a^2 + b^2 + c^2 + 2ab – 2bc – 2ca$ |
| $(a-b+c)^2$ | $a^2 + b^2 + c^2 – 2ab – 2bc + 2ca$ |
| $(a-b-c)^2$ | $a^2 + b^2 + c^2 – 2ab + 2bc – 2ca$ |