### Algebra: Factorization

### Factorization

We have seen how we can factorize an expression with single brackets. Now we will see how we can rewrite an expression with double brackets.

Product sum method

In the example below, we'll explain how we can find the right numbers for the factorization of #x^2+b \cdot x +c#.

Factorize the expression #x^2-6\cdot x-16#.

#x^2-6\cdot x-16=# \((x+2)\cdot(x-8)\)

We are looking for numbers #\purple{m}# and #\purple{n}#, such that the quadratic polynomial #x^2-6\cdot x-16# can be written as #(x+\purple{m})\cdot(x+\purple{n})#. Please note that #\purple{m}# and #\purple{n}# are exchangable. We will expand the brackets and compare the results with the oritginal expression:

\[ x^2+(\purple{m}+\purple{n})\cdot x+\purple{m}\cdot \purple{n} = x^2-6\cdot x-16\tiny\]

Comparison with #x^2-6\cdot x-16# gives \[

\lineqs{\purple{m}+\purple{n} &=& -6\cr \purple{m}\cdot \purple{n} &=& -16}\]If #\purple{m}# and #\purple{n}# are integers, then they are divisors of #-16#. We will check all possible divisors #\purple{m}# with #\purple{m}^2\le 16# (since we then cover all possibilities, because #\purple{m}# and #\purple{n}# van can change roles) and calculate, in each of the cases, the sum of #\purple{m}# and #\purple{n}=\frac{-16}{\purple{m}}#:

\[\begin{array}{|r|c|l|}

\hline

\purple{m}&\purple{n}&\purple{m}+\purple{n}\\

\hline

1&-16&-15\\ \hline -1&16&15\\ \hline 2&-8&-6\\ \hline -2&8&6\\ \hline 4&-4&0\\ \hline -4&4&0 \\

\hline

\end{array}\]

The line of the table with #\purple{m}=2# and #\purple{n}=-8# is the only one with sum #-6#, hence, this is the answer:

\[x^2-6\cdot x-16=(x+2)\cdot(x-8)\tiny.\]

We are looking for numbers #\purple{m}# and #\purple{n}#, such that the quadratic polynomial #x^2-6\cdot x-16# can be written as #(x+\purple{m})\cdot(x+\purple{n})#. Please note that #\purple{m}# and #\purple{n}# are exchangable. We will expand the brackets and compare the results with the oritginal expression:

\[ x^2+(\purple{m}+\purple{n})\cdot x+\purple{m}\cdot \purple{n} = x^2-6\cdot x-16\tiny\]

Comparison with #x^2-6\cdot x-16# gives \[

\lineqs{\purple{m}+\purple{n} &=& -6\cr \purple{m}\cdot \purple{n} &=& -16}\]If #\purple{m}# and #\purple{n}# are integers, then they are divisors of #-16#. We will check all possible divisors #\purple{m}# with #\purple{m}^2\le 16# (since we then cover all possibilities, because #\purple{m}# and #\purple{n}# van can change roles) and calculate, in each of the cases, the sum of #\purple{m}# and #\purple{n}=\frac{-16}{\purple{m}}#:

\[\begin{array}{|r|c|l|}

\hline

\purple{m}&\purple{n}&\purple{m}+\purple{n}\\

\hline

1&-16&-15\\ \hline -1&16&15\\ \hline 2&-8&-6\\ \hline -2&8&6\\ \hline 4&-4&0\\ \hline -4&4&0 \\

\hline

\end{array}\]

The line of the table with #\purple{m}=2# and #\purple{n}=-8# is the only one with sum #-6#, hence, this is the answer:

\[x^2-6\cdot x-16=(x+2)\cdot(x-8)\tiny.\]

The examples below will show that there are also other situations in which we can factorize. We will also see that we can sometimes combine factoring out with factorization in double brackets.

#\left(f-2\right)^2#

As the product of #-2# and #-2# is equal to #4# and the sum to #-4# the following holds:

\[f^2-4\cdot f+4= \left(f-2\right)^2\]

As the product of #-2# and #-2# is equal to #4# and the sum to #-4# the following holds:

\[f^2-4\cdot f+4= \left(f-2\right)^2\]

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