How do you factor #2x^2+x-5#?

2 Answers
Jun 22, 2017

Answer:

#2x^2+x-5=color(red)(2(x+(1+sqrt(41))/4)(x+(1-sqrt(41))/4))#

Explanation:

Apply the quadratic formula
#color(white)("XXX") x=(-b+-sqrt(b^2-4ac))/(2a)#
#color(white)("XXXXXXXXX")# where #ax^2+bx+c=0#

to #2x^2+x-5=0#

obtaining the solutions
#color(white)("XXX") x=(-1+-sqrt(41))/4#
and thus the factors
#color(white)("XXX") (x-(-1+sqrt(41))/4) and (x-(-1-sqrt(41))/4)#

Since
#color(white)("XXX") (x+(1-sqrt(41))/4)(x+(1+sqrt(41))/4)=x^2+x/2-5/2#
we will have an additional constant factor of #2#

Jun 22, 2017

Answer:

#f(x) = 2(x + 1/4 - sqrt41/4)(x + 1/4 + sqrt41/4)#

Explanation:

Use the quadratic function in intercept form (Google Search):
#f(x) = a(x - x_1)(x - x_2).#

#x_1 and x_2# are the #2# real roots of #f(x) = 0#.

They are given by the improved quadratic formula (Socratic Search):

#x = - b/(2a) +- d/(2a)#

In this case:

#D = b^2 - 4ac = 1 + 40 = 41 rarr d = +- sqrt41#
#x = - 1/4 +- sqrt41/4 #
#x_1 = - 1/4 + sqrt41/4#
#x_2 = - 1/4 - sqrt41/4#

#f(x) = a(x - x_1)(x - x_2)#

#f(x) = 2(x + 1/4 - sqrt41/4)(x + 1/4 + sqrt41/4) #