Squaring a Trinomial
The square of a trinomial (a + b + c)2 expands into the sum of the squares of each individual term, plus twice the product of every distinct pair of terms: $$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc $$
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:
$$ = 4a^2 + 9b^2 + 16c^2 - 12ab + 16ac - 24bc $$
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.
