Some equations can be solved by factoring. Factoring is splitting a complex expression into simpler “factors.”

Example Find factors from $$3x+9$$.
Solution  $$3x+9 = 3(x+3)$$
So $$3x+9$$ has been “factored into” 3 and $$x+3$$.
We can say that 3 is the  common factor.

Example Find factors from $$2y+y^2$$.
Solution  $$2y+y^2 = y(2+y)$$
So $$2y+y^2$$ has been “factored into” $$y$$ and $$2+y$$.
We can say that $$y$$ is the common factor.

Example Find factors from $$3(p – q) – r(p – q)$$.
Solution  What is the common factor?

The common factor is $$(p – q)$$.
$$3(p – q) – r(p – q) = (p-q)(3-r)$$
So the equation has been “factored into” $$p-q$$ and $$3-r$$.
We can say that $$p-q$$ is the common factor.

Factor $$4k^2+2k$$.
$$4k^2+2k \\ \, \\ = 2k(2k+1)$$

Factor $$\dfrac{x}{5}+\dfrac{y}{5}$$.
$$\dfrac{x}{5}+\dfrac{y}{5}\\ \, \\ = \dfrac{1}{5} (x+y)$$

Factor $$7(a+b)-b(a+b)$$.
$$7(a+b)-b(a+b)\\ \, \\ = (7-b)(a+b)$$

## Factoring Nearly Identical Terms

Example Find factors from $$x(2 – y) – 5(y – 2)$$.
Solution  Notice that $$(2 – y)$$ in the first term is not the same as the second term $$(y-2)$$.
They are nearly identical, so work around to make them identical!

Simply multiply by (-1)(-1) because  multiplying -1 and -1 produces 1, which does not affect the equation!
$$(y-2) \cdot {\color{red}{(-1)(-1)}} \\ \, \\ = (-1)(-1)(y-2) \\ \, \\ = (-1)(-y+2) \\ \, \\=(-1)(2-y)$$

So we found that $$y – 2 = (-1)(2-y)$$.
Substitute this into the second term in the equation.
$$x(2 – y) – 5\underbrace{(-1)(2-y)}_{\text{Substitution}} \\ \, \\ = x(2-y)+5(2-y)$$

Now we have a common factor of $$(2-y)$$!
$$x(2-y)+5(2-y) = (x+5)(2-y)$$

So the equation has been “factored into” $$x+5$$ and $$2-y$$.
We can say that $$2-y$$ is the common factor.

Factor $$k(2-m)+m(m-2)$$.
From the second term $$(m-2)$$, we multiply it by $$(-1)(-1)$$ and we get $$(-1)(2-m)$$.
$$(m-2) = (-1)(-1)(m-2) = (-1)(2-m)$$

Substitute this value into the second term.
$$k(2-m) + m\underbrace{(-1)(2-m)}_{\text{Substitution}} \\ \, \\ = k(2-m) – m(2-m) \\ \, \\ = (k-m)(2-m)$$

Factor $$8(1-a)+2(2a-2)$$.
First, factor $$(2a-2)$$.
$$(2a-2)=2(a-2)$$

Then we have
$$8(1-a)+2(2a-2) \\ = 8(1-a)+(2)(2)(a-1) \\ = 8(1-a)+4(a-1)$$

From the second term $$(a-1)$$, we multiply it by $$(-1)(-1)$$ and we get $$(-1)(1-a)$$.
$$(a-1) \\ = {\color{red}{(-1)(-1)}}(a-1) \\ = (-1)(1-a)$$

Substitute this value into the second term.
$$8(1-a) + 4\underbrace{(-1)(1-a)}_{\text{Substitution}} \\ \, \\ = 8(1-a) -4(1-a) \\ \, \\ = (8-4)(1-a) \\ \, \\ = 4(1-a)$$