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

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

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


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 \(k(2-m)+m(m-2)\).
Answer
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 \(8(1-a)+2(2a-2)\).
Answer
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.
\(8(1-a) + 4\underbrace{(-1)(1-a)}_{\text{Substitution}} \\ \, \\ = 8(1-a) -4(1-a) \\ \, \\ = (8-4)(1-a) \\ \, \\ = 4(1-a) \)

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