How do you factor #7y^2-11y-27#?

2 Answers
Jan 9, 2017

Answer:

#7y^2-11y-27 = 1/28(14y-11-sqrt(877))(14y-11+sqrt(877))#

Explanation:

The difference of squares identity can be written:

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

Use this with #a=(14y-11)# and #b=sqrt(877)#

First premultiply by #7*2^2 = 28#, then divide by it after completing the square:

#28(7y^2-11y-27) = 196y^2-308y-756#

#color(white)(28(7y^2-11y-27)) = (14y)^2-2(14y)(11)+121-877#

#color(white)(28(7y^2-11y-27)) = (14y-11)^2-(sqrt(877))^2#

#color(white)(28(7y^2-11y-27)) = ((14y-11)-sqrt(877))((14y-11)+sqrt(877))#

#color(white)(28(7y^2-11y-27)) = (14y-11-sqrt(877))(14y-11+sqrt(877))#

Hence:

#7y^2-11y-27 = 1/28(14y-11-sqrt(877))(14y-11+sqrt(877))#

Jan 9, 2017

Answer:

(1/28)(14x - 11 - sqrt877)(14x - 11 + sqrt877)

Explanation:

There is another classical way.
f(x) = a(x - x1)(x - x2),
where x1 and x2 are the 2 real roots of the quadratic equation
f(x) = 7x^2 - 11x - 27 = 0.
#D = d^2 = b^2 - 4ac = 121 + 756 = 877# --> #d = +- sqrt877#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 11/14 +- sqrt877/14#
#x1 = 11/14 + sqrt877/14#
#x2 = 11/14 - sqrt877/14#
The factored form will be:
#f(x) = (7)(x - 11/14 - sqrt877/14)(x - 11/14 - sqrt877/14) = #
#= 1/28(14x - 11 - sqrt877)(14x - 11 + sqrt877)#