How do you factor the expression #x^2-x-36#?

1 Answer
Jan 5, 2016

Answer:

Use the quadratic formula to find:

#x^2-x-36 = (x-(1-sqrt(145))/2)(x-(1+sqrt(145))/2)#

Explanation:

You would like to find a pair of factors of #36# that differ by #1#. Unfortunately, no such pair exists, so this polynomial will not factor with integer coefficients.

Let's check the discriminant: #x^2-x-36# is in the form #ax+bx+c# with #a=1#, #b=-1# and #c=-36#. This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-1)^2-(4*1*-36) = 1+144 = 145#

Since this is positive but not a perfect square, the quadratic has factors with Real irrational coefficients.

We can find these factors using the quadratic formula. The zeros of the quadratic are given by:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)#

#=(1+-sqrt(145))/2#

That is #x_1 = (1-sqrt(145))/2# and #x_2 = (1+sqrt(145))/2#

So:

#x^2-x-36 = (x-x_1)(x-x_2) = (x-(1-sqrt(145))/2)(x-(1+sqrt(145))/2)#