How do you factor #x^{2}-16x-96#?

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Answer 1 · 2018-04-03T08:10:56

#(x-\frac{16 + 8sqrt(10)}{2})(x-\frac{16 -8sqrt(10)}{2})#

You can use the standard formula

#x_{1,2} = \frac{-b \pm \sqrt(b^2-4ac)}{2a}#

if the equation is

#ax^2+bx+c=0#

to find its roots. In your case, we have

#x_{1,2} = \frac{16 \pm \sqrt(256-4*1*96)}{2*1}#

#=\frac{16 \pm \sqrt(256+384)}{2}#

#=\frac{16 \pm \sqrt(640)}{2}#

As a side note, we can write

#\sqrt(640) = \sqrt(64*10) = \sqrt(64)*\sqrt(10) = 8sqrt(10)#

So, the solutions become

#x_{1,2} = \frac{16 \pm 8sqrt(10)}{2}#

So, we can factor the polynomial as

#(x-\frac{16 + 8sqrt(10)}{2})(x-\frac{16 -8sqrt(10)}{2})#