### 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-45# as a product of linear factors.

#x^2+4\cdot x-45=# \((x-5)\cdot(x+9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+4\cdot x-45# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2+4\cdot x-45\tiny\]

A comparison with #x^2+4\cdot x-45# gives \[

\lineqs{p+q &=& -4\cr p\cdot q &=& -45}\] If #p# and #q# are integers, they are divisors of #-45#. We go through all possible divisors #p# with #p^2\le |-45|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-45}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-45&-44\\ \hline -1&45&44\\ \hline 3&-15&-12\\ \hline -3&15&12\\ \hline 5&-9&-4\\ \hline -5&9&4 \\

\hline

\end{array}\]

The line of the table with #p=5# and #q=-9# is the only one with sum #-4#, hence, this is the answer:

\[x^2+4\cdot x-45=(x-5)\cdot(x+9)\tiny.\]

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+4\cdot x-45# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2+4\cdot x-45\tiny\]

A comparison with #x^2+4\cdot x-45# gives \[

\lineqs{p+q &=& -4\cr p\cdot q &=& -45}\] If #p# and #q# are integers, they are divisors of #-45#. We go through all possible divisors #p# with #p^2\le |-45|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{-45}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&-45&-44\\ \hline -1&45&44\\ \hline 3&-15&-12\\ \hline -3&15&12\\ \hline 5&-9&-4\\ \hline -5&9&4 \\

\hline

\end{array}\]

The line of the table with #p=5# and #q=-9# is the only one with sum #-4#, hence, this is the answer:

\[x^2+4\cdot x-45=(x-5)\cdot(x+9)\tiny.\]

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