### Functions: Quadratic functions

### Factorization

The *quadratic formula* can always be applied to a quadratic equation, but it is certainly not always the fastest way. Sometimes you can use factorization.

Write the expression #x^2-4\cdot x-12# as a product of linear factors.

Solution #x^2-4\cdot x-12=# \((x+2)\cdot(x-6)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-4\cdot x-12# can be written as #(x-p)\cdot(x-q)#. If the absolute value of #p# is greater than #q#, we interchange them, so #|p|\le |q|#. We expand the brackets and compare the result with the original expression:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-4\cdot x-12\tiny\]

A comparison with #x^2-4\cdot x-12# gives \[

\lineqs{p+q &=& 4\cr p\cdot q &=& -12}\] If #p# and #q# are integers, they are divisors of #-12#. We go through all possible divisors #p# with #p^2\le |-12|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-12}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-12&-11\\ \hline -1&12&11\\ \hline 2&-6&-4\\ \hline -2&6&4\\ \hline 3&-4&-1\\ \hline -3&4&1 \\

\hline

\end{array}\]

The line of the table with #p=-2# and #q=6# is the only one with sum #4#, hence, this is the answer:

\[x^2-4\cdot x-12=(x+2)\cdot(x-6)\tiny.\]

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-4\cdot x-12# can be written as #(x-p)\cdot(x-q)#. If the absolute value of #p# is greater than #q#, we interchange them, so #|p|\le |q|#. We expand the brackets and compare the result with the original expression:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-4\cdot x-12\tiny\]

A comparison with #x^2-4\cdot x-12# gives \[

\lineqs{p+q &=& 4\cr p\cdot q &=& -12}\] If #p# and #q# are integers, they are divisors of #-12#. We go through all possible divisors #p# with #p^2\le |-12|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-12}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-12&-11\\ \hline -1&12&11\\ \hline 2&-6&-4\\ \hline -2&6&4\\ \hline 3&-4&-1\\ \hline -3&4&1 \\

\hline

\end{array}\]

The line of the table with #p=-2# and #q=6# is the only one with sum #4#, hence, this is the answer:

\[x^2-4\cdot x-12=(x+2)\cdot(x-6)\tiny.\]

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