### 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^2-15\cdot x+56# as a product of linear factors.

#x^2-15\cdot x+56=# \((x-7)\cdot(x-8)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-15\cdot x+56# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-15\cdot x+56\tiny\]

A comparison with #x^2-15\cdot x+56# gives \[

\lineqs{p+q &=& 15\cr p\cdot q &=& 56}\] If #p# and #q# are integers, they are divisors of #56#. We go through all possible divisors #p# with #p^2\le |56|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{56}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&56&57\\ \hline -1&-56&-57\\ \hline 2&28&30\\ \hline -2&-28&-30\\ \hline 4&14&18\\ \hline -4&-14&-18\\ \hline 7&8&15\\ \hline -7&-8&-15 \\

\hline

\end{array}\]

The line of the table with #p=7# and #q=8# is the only one with sum #15#, hence, this is the answer:

\[x^2-15\cdot x+56=(x-7)\cdot(x-8)\tiny.\]

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2-15\cdot x+56# 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:

\[ x^2-(p+q)\cdot x+p\cdot q = x^2-15\cdot x+56\tiny\]

A comparison with #x^2-15\cdot x+56# gives \[

\lineqs{p+q &=& 15\cr p\cdot q &=& 56}\] If #p# and #q# are integers, they are divisors of #56#. We go through all possible divisors #p# with #p^2\le |56|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{56}{p}#:

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

\hline

p&q&{p+q}\\

\hline

1&56&57\\ \hline -1&-56&-57\\ \hline 2&28&30\\ \hline -2&-28&-30\\ \hline 4&14&18\\ \hline -4&-14&-18\\ \hline 7&8&15\\ \hline -7&-8&-15 \\

\hline

\end{array}\]

The line of the table with #p=7# and #q=8# is the only one with sum #15#, hence, this is the answer:

\[x^2-15\cdot x+56=(x-7)\cdot(x-8)\tiny.\]

Unlock full access

Teacher access

Request a demo account. We will help you get started with our digital learning environment.

Student access

Is your university not a partner?
Get access to our courses via

Or visit omptest.org if jou are taking an OMPT exam.

**Pass Your Math**independent of your university. See pricing and more.Or visit omptest.org if jou are taking an OMPT exam.