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

2 Answers
Jan 9, 2017

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

(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)