How do you factor #t^ { 2} + 5t - 28#?

1 Answer
George C.
Nov 20, 2017

#t^2+5t-28 = (t+5/2-sqrt(137)/2)(t+5/2+sqrt(137)/2)#

Explanation:

Complete the square, then use the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A=(2t+5)# and #B=sqrt(137)# as follows:

#4(t^2+5t-28) = 4t^2+20t-112#

#color(white)(4(t^2+5t-28)) = (2t^2)+2(2t)(5)+5^2-137#

#color(white)(4(t^2+5t-28)) = (2t+5)^2-(sqrt(137))^2#

#color(white)(4(t^2+5t-28)) = ((2t+5)-sqrt(137))((2t+5)+sqrt(137))#

#color(white)(4(t^2+5t-28)) = (2t+5-sqrt(137))(2t+5+sqrt(137))#

#color(white)(4(t^2+5t-28)) = 4(t+5/2-sqrt(137)/2)(t+5/2+sqrt(137)/2)#

So:

#t^2+5t-28 = (t+5/2-sqrt(137)/2)(t+5/2+sqrt(137)/2)#

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