How do you factor the expression #x^2-20x+48#?

2 Answers
Apr 4, 2016

not factorable

Explanation:

#D = b^2 - 4ac = 400 - 192 = 208#
Since D is not a perfect square, this trinomial can't be factored.

Apr 4, 2016

This expression has no rational factors.
The irrational factors are #(x-10-2sqrt(13))*(x-10+2sqrt(13))#

Explanation:

As shown by a previous answer, this expression has no rational factors. If, however, we allow irrationals, we can factor the expression using its zeros. Graphing the expression we see that it has two real zeros:

graph{x^2-20*x+48 [-2, 22, -56, 20]}

Where the factors will be in the form:

#0=(x-a_+)*(x-a_-)#

Where #a_+# and #a_-# are the zeros of the expression. Setting the expression equal to zero:

#0=x^2-20x+48#

And then using the quadratic equation:

#a_+-=(20+-sqrt(400-4*48))/2=10+-2sqrt(13)#

Therefore, the expression factors to:

#(x-10-2sqrt(13))*(x-10+2sqrt(13))#