How do you factor #\sqrt { 5} x ^ { 2} + 2x - 3\sqrt { 5} #?

1 Answer
George C.
Aug 4, 2017

#sqrt(5)x^2+2x-3sqrt(5) = (sqrt(5)x-3)(x+sqrt(5))#

Explanation:

We can complete the square and use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x+1/sqrt(5))# and #b=4/sqrt(5)# as follows:

#sqrt(5)x^2+2x-3sqrt(5) = sqrt(5)(x^2+2/sqrt(5)x-3)#

#color(white)(sqrt(5)x^2+2x-3sqrt(5)) = sqrt(5)(x^2+2(1/sqrt(5))x+1/5-16/5)#

#color(white)(sqrt(5)x^2+2x-3sqrt(5)) = sqrt(5)((x+1/sqrt(5))^2-(4/sqrt(5))^2)#

#color(white)(sqrt(5)x^2+2x-3sqrt(5)) = sqrt(5)((x+1/sqrt(5))-4/sqrt(5))((x+1/sqrt(5))+4/sqrt(5))#

#color(white)(sqrt(5)x^2+2x-3sqrt(5)) = sqrt(5)(x-3/sqrt(5))(x+5/sqrt(5))#

#color(white)(sqrt(5)x^2+2x-3sqrt(5)) = (sqrt(5)x-3)(x+sqrt(5))#

Related questions
Impact of this question
1216 views around the world
You can reuse this answer
Creative Commons License