How do you solve # (x+5)(x^2-7x+12)=0#?

1 Answer
KillerBunny
Sep 28, 2015

The solutions are #-5#, #3# and #4#.

Explanation:

A multiplications gives zero as a result if and only if at least one of his factors equals zero. So that's what you need to impose.

The first factor is quite simple: #x+5# equals zero if and only if #x=-5#.

The second factor is a parabola, whose zeroes you may find through the classical formula #{-b \pm \sqrt{b^2-4ac}/2a#, but in simple cases as this, I prefer this simplier one: if the coefficient of #x^2# is one, then you can read your equation like this:

#x^2-sx+p=0#, where #s# is the sum of the roots, and #p# is their product. So, you have #-s=-7#, and #p=12#. This means that we are looking for two numbers #a# and #b# such that #a+b=7#, and #a*b=12#. It's easy to see, with barely no calculations, that these numbers are #3# and #4#.

So, your equation is solved by three numbers, namely #-5#, #3# and #4#.

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