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 \(4k^2+2k\).

Factor \(\dfrac{x}{5}+\dfrac{y}{5}\).

Factor \(7(a+b)-b(a+b)\).

Factor \(\dfrac{x}{5}+\dfrac{y}{5}\).

Factor \(7(a+b)-b(a+b)\).

**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 \(k(2-m)+m(m-2)\).

Factor \(8(1-a)+2(2a-2)\).

Factor \(8(1-a)+2(2a-2)\).

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