### 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^2-3\cdot x-4# as a product of linear factors.

#x^2-3\cdot x-4=# \((x+1)\cdot(x-4)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-3\cdot x-4# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-3\cdot x-4\tiny\]

A comparison with #x^2-3\cdot x-4# gives \[

\lineqs{p+q &=& 3\cr p\cdot q &=& -4}\] If #p# and #q# are integers, they are divisors of #-4#. We go through all possible divisors #p# with #p^2\le |-4|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-4}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-4&-3\\ \hline -1&4&3\\ \hline 2&-2&0\\ \hline -2&2&0 \\

\hline

\end{array}\]

The line of the table with #p=-1# and #q=4# is the only one with sum #3#, hence, this is the answer:

\[x^2-3\cdot x-4=(x+1)\cdot(x-4)\tiny.\]

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-3\cdot x-4# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-3\cdot x-4\tiny\]

A comparison with #x^2-3\cdot x-4# gives \[

\lineqs{p+q &=& 3\cr p\cdot q &=& -4}\] If #p# and #q# are integers, they are divisors of #-4#. We go through all possible divisors #p# with #p^2\le |-4|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-4}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-4&-3\\ \hline -1&4&3\\ \hline 2&-2&0\\ \hline -2&2&0 \\

\hline

\end{array}\]

The line of the table with #p=-1# and #q=4# is the only one with sum #3#, hence, this is the answer:

\[x^2-3\cdot x-4=(x+1)\cdot(x-4)\tiny.\]

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