What are all the possible rational zeros for #f(x)=x^2+8x+10#?

1 Answer
Mar 28, 2018

Answer:

The roots are #-4+\sqrt{6}# and #-4-\sqrt{6}#.

Explanation:

You start with #f(x)=x^2+8x+10#. First you have to establish its degree, which is #2# because the highest exponent is #2#. That being said, you can apply the quadratic formula, namely:

#ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#

In this particular case, #a# is #1#, #b# is #8# and #c# is #10#, and therefore:

#x=\frac{-8\pm\sqrt{8^2-4\cdot 1\cdot 10}}{2\cdot 1}#

#x=\frac{-8\pm\sqrt{64-40}}{2}#

#x=\frac{-8\pm\sqrt{24}}{2}=\frac{-8\pm\sqrt{2\cdot 2\cdot 6}}{2}#

#x=\frac{-8\pm2\sqrt{6}}{2}=-4\pm\sqrt{6}#

Hence, the two roots are #-4+\sqrt{6}# and #-4-\sqrt{6}#.