Squaring a Trinomial

Squaring a trinomial like (a + b + c)² means taking the sum of the squares of each term and adding twice the product of every pair of distinct terms. $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$

The terms $a$, $b$, and $c$ may be monomials or even more complex polynomials.

This identity can be extended to polynomials with more than three terms.

$$\left(\sum_{i=1}^{n} x_i\right)^2 = \sum_{i=1}^{n} x_i^2 + 2 \sum_{1 \le i < j \le n} x_i x_j$$

Simply square each term individually and add twice the product of every distinct pair of terms.

For instance, to square a four-term polynomial (a + b + c + d)², write: $$ (a + b + c + d)^2 = $$ $$ = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd $$

A Practical Example

Let’s look at a specific case:

$$ (2a - 3b + 4c)^2 $$

Applying the general formula:

$$ (2a - 3b + 4c)^2 = (2a)^2 + (-3b)^2 + (4c)^2 + 2 \cdot (2a) \cdot (-3b) + 2 \cdot (2a) \cdot (4c) + 2 \cdot (-3b) \cdot (4c) $$

Now simplify each term:

$$ = 4a^2 + 9b^2 + 16c^2 - 12ab + 16ac - 24bc $$

Mind the Signs in the Cross Terms! When squaring a trinomial - or any polynomial, really - it’s easy to overlook one crucial detail: the signs. Remember: Squaring a negative always gives a positive result. That’s because $( -x )^2 = x^2$. But cross terms? That’s where signs matter. They don’t disappear - and they can completely change the outcome. Take this example: $$ (a - b)^2 = a^2 - 2ab + b^2 $$ The middle term comes from $2 \cdot a \cdot (-b)$, which equals $-2ab$. So, the minus sign stays - and makes a difference. In short: when expanding, keep an eye on the signs. They matter more than you think!

Algebraic Proof

Let’s derive the identity from scratch.

Start with the square of a trinomial:

$$ (a + b + c)^2 $$

Group terms using the associative property of addition:

$$ [(a + b) + c]^2 $$

This is now the square of a binomial, where $ (a + b) $ acts as one term and $ c $ as the other.

Apply the standard identity for squaring a binomial, $ (x + y)^2 = x^2 + 2xy + y^2 $:

$$ [(a + b) + c]^2 = (a + b)^2 + 2(a + b)c + c^2 $$

Expand the expression:

$$ = a^2 + 2ab + b^2 + 2ac + 2bc + c^2 $$

Rewriting in a more natural order using the commutative property:

$$ = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$

Which is exactly the identity we aimed to prove.

Geometric Interpretation

We can also understand this identity through geometry.

Take three segments with lengths $a$, $b$, and $c$.

Place them end to end to form one side of a square whose length is $a + b + c$.

The area of the square is the product of its side with itself: $ (a + b + c) \cdot (a + b + c) = (a + b + c)^2 $.

We can also divide this square into smaller squares and rectangles whose areas we can calculate separately.

Adding up those individual areas gives:

$$ a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$

Which confirms the identity:

$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$

Additional Notes

A few helpful remarks and alternative approaches:

• Alternative Method
  Since the square of a trinomial appears less frequently than the binomial case - especially in secondary-level math - it’s easy to forget the formula. In such cases, you can rewrite the expression $ (a + b + c)^2 $ as a square of a binomial by grouping: $$ (a + b + c)^2 = ((a + b) + c)^2 $$ Now apply the binomial square identity $ (A + B)^2 = A^2 + 2AB + B^2 $, where A = (a + b) and B = c: $$ = (a + b)^2 + 2(a + b)c + c^2 $$ Expand each part step by step: $$ = a^2 + 2ab + b^2 + 2ac + 2bc + c^2 $$ Rearranged in standard form: $$ = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$ As expected, we arrive at the same result.

And so on.