Squaring a Quadrinomial

To square a quadrinomial of the form $ a + b + c + d $, we apply the formula for the square of the sum of four terms:

$$ (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + \\ + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd $$

Where:

$ a^2, b^2, c^2, $ and $ d^2 $ are the squares of each individual term.
$ 2ab, 2ac, 2ad, 2bc, 2bd, $ and $ 2cd $ represent the cross terms—each one being twice the product of a distinct pair of terms.

Note. An alternative method involves rewriting the quadrinomial as the sum of two binomials and applying the identity (A + B)² = A² + 2AB + B². For example, let A = (a + b) and B = (c + d): $$ (a + b + c + d)^2 = ((a + b) + (c + d))^2 $$ $$ = A^2 + 2AB + B^2 $$ Expanding this: $$ = (a + b)^2 + 2(a + b)(c + d) + (c + d)^2 $$ This leads to the same result as the general formula.

Worked Example
Proof

Worked Example

Let’s consider the quadrinomial $ 2 + 3x + y + 5z $. To square it, we follow the general procedure:

$$ (2 + 3x + y + 5z)^2 $$

We compute the square using the formula:

$$ = 2^2 + (3x)^2 + y^2 + (5z)^2 + 2(2)(3x) + 2(2)(y) + 2(2)(5z) + 2(3x)(y) + 2(3x)(5z) + 2(y)(5z) $$

Simplifying each term:

$$ = 4 + 9x^2 + y^2 + 25z^2 + 12x + 4y + 20z + 6xy + 30xz + 10yz $$

This is the fully expanded form of the square of the given quadrinomial.

Proof

To derive the general formula, we start with the square of a quadrinomial:

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

Which is the product of the expression with itself:

$$ (a + b + c + d)(a + b + c + d) $$

Using distributive multiplication (the distributive property of multiplication over addition), we expand term by term:

$$ a(a + b + c + d) + b(a + b + c + d) + c(a + b + c + d) + d(a + b + c + d) $$

Which yields:

$$ a^2 + ab + ac + ad + ab + b^2 + bc + bd + ac + bc + c^2 + cd + ad + bd + cd + d^2 $$

Now combining like terms:

$$ a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd $$

And this confirms the expansion of the square of a four-term sum.