How do you solve #x^2-4x-5=0# by factoring?

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Answer 1 · 2015-09-29T12:31:27

The solutions are
# color(blue)(x=-1#
# color(blue)(x=+5#

#x^2−4x−5=0#

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

#N_1*N_2 = a*c = 1*(-5)= -5#
AND
#N_1 +N_2 = b = -4#

After trying out a few numbers we get #N_1 = -5# and #N_2 =1#

#(1)*(-5) = -5#, and #1+(-5)= -4#

#x^2−color(blue)(4x)−5=x^2−color(blue)(5x+1x)−5#

#=x(x-5) +1(x−5)#

#=color(blue)((x+1)(x-5) # are the factors.

We can now obtain the solutions by equating the factors to zero.
#=x+1 =0, color(blue)(x=-1#
#=x-5 =0, color(blue)(x=+5#