Question

How do you factor `5x ^ { 2} + 2x - 7- 11x ^ { 2}`?

Answer 1

`f(x) = - 6x^2 + 2x - 7`
`D = b^2 - 4ac = 4 - 168 = - 164`
Since D< 0 and is not a perfect square, then, this trinomial can't be factored.

Answer 2

Factors are `f(x)=5(x-1/5-isqrt41/5)(x-1/5+isqrt41/5)`

Explanation:

As `f(x)=5x^2+2x-7-11x^2=-6x^2+2x-7`

and discrimant `Delta=b^2-4ac=2^2-4xx(-6)xx(-7)=4-168=-164` is negative

we can have factors with only complex coefficients.

Quadratic formula gives zeros as

`(-2+-sqrtDelta)/(2xx(-5))=(-2+-sqrt(-164))/(-10)`

= `(-2+-2isqrt41)/(-10)`

i.e. `1/5+isqrt41/5` or `1/5-isqrt41/5`

and factors are `f(x)=5(x-1/5-isqrt41/5)(x-1/5+isqrt41/5)`