How do you factor #x^2-8x-15# ?

The Completing the Square Method basically "forces" the existence of a perfect square trinomial in order to easily factor an equation where factoring by grouping is impossible.

For #x^2-8x-15#, the first step would be to make the equation equal to zero and add 15 to both sides.

#x^2-8x=15#

Next, we need to turn the binomial on the left side of the equation into a perfect trinomial. We can do this by dividing the coefficient of the "middle x-term", which would be #-8#, in half then squaring it.

#(-8)/2=-4# and #(-4)^2=16#

We then add the result to both sides of the equation.

#x^2-8x+16=15+16#

Now, we can factor the perfect square trinomial and simplify #15+16#.

#(x+4)^2=31#

Now, we subtract #31# from both sides of the equation.

#(x+4)^2-31#

In order to factor these terms, both of them need to be the "squared version of their square rooted form" so that the terms stay the same when factoring.

#(x+4)^2-sqrt(31)^2#

Now, we can factor them. Since the expression follows the case #a^2-b^2#, where, in this case, #a=(x+4)# and #b=sqrt(31)#, we can factor them by following the Difference of Squares Formula: #a^2-b^2=(a+b)(a-b)#.

Your final answer would be:
#((x+4)+sqrt(31))((x+4)-sqrt(31))#

In the quadratic polynomial #ax^2+bx+c#, where #a#, #b# and #c# are rational numbers, if #b^2-4ac# is positive but not the square of a rational number, roots are irrational, in such cases it may not be possible to factorize the polynomial. In such cases factorization, we may use quadratic formula #(-b+--sqrt(b^2-4ac))/(2a)# to find zeros, and if they are #p# and #q#, the factors are #a(x-p)(x-q)#.

In #x^2-8x-15#, #b^2-4ac=(-8)^2-4xx1xx(-15)=64+60=124# is not the square of a rational number, and using quadratic formula roots are

#(-(-8)+--sqrt((-8)^2-4xx1xx(-15)))/(2xx1)#

= #(8+-sqrt124)/2=(8+-2sqrt31)/2=4+-sqrt31#, and factors are

#x^2-8x-15=(x-4-sqrt31)(x-4+sqrt31)#