How do you factor #49p^2+28ps+4s^2#?

1 Answer
Dec 4, 2016

#49p^2+28ps+4s^2=(7p+2s)^2#

Explanation:

The equation #49p^2+28ps+4s^2# is in fact a quadratic equation. Observe that it has two variables and each monomial has a degree of #2#. Such a polynomial is called homogeneous.

If we divide each term of the polynomial by #s^2#, we get #49p^2/s^2+28(ps)/s^2+4# or #49(p/s)^2+28p/s+4# and putting #p/s=x#, we get

#49x^2+28x+4#.

Now we can factorize it like any quadratic equation by either completing square method or splitting the middle term.

What is typical about this is that in fact polynomial is a complete square.

As #49p^2+28ps+4s^2#

= #(7p)^2+2xx7pxx2s+(2s)^2#

= #(7p+2s)^2#

hence the factors.