Question

How do you factor `q^2 - 7q - 10`?

Answer 1

`x_1= 7/2+1/2*(sqrt(89))`
`x_2= 7/2-1/2*(sqrt(89))`

Explanation:

quadratic formula
`x_1,_2 =(-b+- sqrt(b^2-4ac))/(2a)`
here `a=1,b=-7,c=-10`
so
`x_1 =(-(-7)+ sqrt((-7)^2-4*1*(-10))/(2*1))`
`= 7/2+1/2*sqrt(89)`
`x_2 =(-(-7)- sqrt((-7)^2-4*1*(-10))/(2*1))`
`= 7/2-1/2*sqrt(89)`

Answer 2

This cannot be factored using whole numbers. (It can be factored using some square roots.)

Explanation:

If we are to factor `q^2-7q-10` using whole numbers, we would need to have `(q +- "some wholenumber")(q +- "some other number")`

Because we would use FOIL to multiply these two expression, we would need the two numbers (they will be the Last) to have a product of `-10`.

the Outside + Inside need to be `-7q`, so we needthe larger of the 'lasts' to be negative and the smaller to be positive.

The only possibilities are:

`(q+1)(q-10)` (which does not have `-7q` in the middle -- it has `-9q`)

`(q+2)(q-5)` (which does not have `-7q` in the middle -- it has `-3q`)

Since the only two ways to get `q^2` First`and`-10# Last with a negative in the middle don't work. Nothing will work.

If you have learned how to solve `q^2-7q-10 = 0` using square roots, you can factor the expression using those solutions.

The solutions are `(7+sqrt89)/2` and `(7-sqrt89)/2` and the factors are:

`(x-(7+sqrt89)/2)(x-(7-sqrt89)/2)`