Partial Factoring

Partial factoring is a technique used in the factoring and decomposition of polynomials. It applies when only certain terms within a polynomial share a common factor - either a number or a variable. In such cases, the shared factor can be factored out from those specific terms: $$ ac + ad + bc = a \cdot (c + d) + bc $$

Like complete factoring, partial factoring is based on the distributive property of multiplication over addition.

The key difference is that in partial factoring, the common factor is extracted from only a portion of the polynomial - not all of it.

In some cases, performing a series of partial factorizations may eventually lead to a complete factorization of the entire expression.

$$ (a + b)(c + d) $$ $$ a(c + d) + b(c + d) $$ $$ ac + ad + bc + bd $$

Note. The steps involved in partial or complete factoring are essentially the reverse of the steps used in multiplying two polynomials: $$ (a + b)(c + d) $$ $$ a(c + d) + b(c + d) $$ $$ ac + ad + bc + bd $$

A Practical Example

Example 1

Consider the polynomial:

$$ 3ab - 4bc + cd $$

The first two terms, 3ab and - 4bc, share a common factor: the variable b.

$$ 3a \color{red}b - 4 \color{red}bc + cd $$

We can factor out b from those two terms:

$$ b \cdot (3a - 4c) + cd $$

This is a partial factorization, as it involves only part of the expression.

Example 2

Now take the polynomial:

$$ 6ac - 12bd + 9bc - 8ad $$

The terms 6ac and - 8ad both contain the variable a as a common factor:

$$ 6\color{red}ac - 12bd + 9bc - 8\color{red}ad $$

We begin by factoring out a from these two terms:

$$ a \cdot (6c - 8d) - 12bd + 9bc $$

Next, we observe that the binomial (6c - 8d) can be simplified, since both terms are divisible by 2.

So we factor out the 2:

$$ a \cdot (6c - 8d) - 12bd + 9bc $$

$$ a \cdot (\color{red}2 \cdot 3c - \color{red}2 \cdot 4d) - 12bd + 9bc $$

$$ 2a \cdot (3c - 4d) - 12bd + 9bc $$

Now, let’s look at the next two terms: - 12bd and 9bc. They both contain b as a common factor:

$$ 2a \cdot (3c - 4d) - 12\color{red}bd + 9\color{red}bc $$

We perform another partial factorization, this time factoring out b:

$$ 2a \cdot (3c - 4d) + b \cdot (9c - 12d) $$

Now examine the binomial (9c - 12d). Since both terms are divisible by 3, we can simplify further:

$$ 2a \cdot (3c - 4d) + b \cdot (9c - 12d) $$

$$ 2a \cdot (3c - 4d) + b \cdot (\color{red}3 \cdot 3c - \color{red}3 \cdot 4d) $$

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

At this point, we notice that both terms share the binomial (3c - 4d). This allows us to perform a final complete factorization:

$$ 2a \cdot \color{red}{(3c - 4d)} + 3b \cdot \color{red}{(3c - 4d)} $$

Factoring out (3c - 4d) gives us:

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

This is the fully factored, irreducible form of the original polynomial.

Note. In total, we performed four partial factorizations followed by a complete factorization to express the polynomial in its simplest form.

And so on.