Question

How do you factor `60x ^ { 2} + 33x - 9= 0`?

Answer 1

# (15x - 3)(4x +3)

Explanation:

In the form`Ax^2 + Bx + C`

The C is negative. so one factor must be positive and one must be negative.

The B value is positive so the positive factors must be larger than the negative factors.

possible factors for 9 are `9 xx 1 and 3 xx 3`

possible factors for 60 are 20 x 3
15 x 4
12 x 5
10 x 6
`3 xx 15 - 4 xx 3 = 33` so the factors must be

`15 xx 3` for 60 and `3 xx 3` for 9 so

( 15x -3 ) xx ( 4x + 3)

Answer 2

`60x^2+33x-9=3(5x-1)(4x+3)=0`

Explanation:

To factorize `60x^2+33x-9=0`, we shuld split the middle term `+33`, the coefficient of `x`, in two parts whose product is product of the coefficient of `x^2` and constant term i.e. `60xx(-9)=-540`.

As the product is negative and sum is positive the two parts must be of opposite sign and larger nuumber would be positive. Factors of `540` are

`1,540),(2,270),(3,180),(4,135),(5,108),(6,90),(9,60),(10,54),(12,45).....`

Hence the two numbers are `45`and `-12`

and `60x^2+33x-9`

= `60x^2+45x-12x-9`

= `15x(4x+3)-3(4x+3)`

= `(15x-3)(4x+3)`

= `3(5x-1)(4x+3)`