Question

How do you factor the trinomial `x^2+3x-36=0`?

Answer 1

`x^2+3x-36=0` can be solved by quadratic formula but cannot be solved by integer factors. In case, you are looking for a solution it is explained below.

Explanation:

Quadratic equations of for `ax^2+bx+c=0` can be solved by the quadratic formula.

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Our question `x^2+3x-36 =0`
Compare with `ax^2+bx+c=0`

`a=1`, `b=3` and `c=-36`

Substitute these values of `a`,`b` and `c` in the formula

`x =(-(3)+-sqrt((3)^2-4(1)(-36)))/(2(1))`

`x=(-3+-sqrt(9+ 144))/2`

`x=(-3+-sqrt(153))/2`

`x=(-3+-sqrt(3^2*17))/2`
`x=(-3+-3sqrt(17))/2`

Solutions:

`x=(-3-3sqrt(17))/2` or `x=(-3+3sqrt(17))/2`

Real (non-integer) Factors
`x^2+3x-36 = (x+(3-3sqrt(17))/2)(x+(3+3sqrt(17))/2)`