Question

How do you factor `9-2x=3x^2`?

Answer 1

It doesn't factor

Explanation:

rearrange by adding the `2x` and subtracting the 9

`3x^2+2x-9=0`

You cannot factor it but you can solve it using the formula and get

`x=[-2\pmsqrt112]/6`

`x=[-1\pm2sqrt7]/3`

#x=1.4305008 or x=-2.0971675

Answer 2

`color(blue)((-3x-1-2sqrt(7))(x+(1-sqrt(7))/(3))`

Explanation:

The quadratic has been given as an equation, and not as an expression, which would imply a solution is sought rather than a general factorization.

If we express this as:

`-3x^3-2x+9`

We can factor this, but not using the general method. We would have to find the roots of `-3x^3-2x+9=0` and work in reverse.

We first recognize that if:

`alpha and beta` are the roots to a quadratic then:

`a(x-alpha)(x-beta)` are the factors of this quadratic. where `a` is a multiplier.

Roots using quadratic formula:

`x=(-(-2)+-sqrt((-2)^2-4(-3)(9)))/((2)(-3))`

`x=(1+-2sqrt(7))/(-3)`

`x=(-1+2sqrt(7))/(3)`

`x=(-1-2sqrt(7))/(3)`

`:.`

`a(x-((-1+2sqrt(7))/(3)))(x-((-1-2sqrt(7))/(3)))`

`a(x+(1-2sqrt(7))/(3))(x+(1+2sqrt(7))/(3))`

`a=-3`

`(-3x-1-2sqrt(7))(x+(1-sqrt(7))/(3))`

The value of the multiplier `bba` is the coefficient of `x^2` in the original expression. The reason we need this in:

`a(x-alpha)(x-beta)`

is because, when we solve for the roots using the quadratic formula or completion of the square ( these are one and the same thing ), we divide by the coefficient of `x^2`, so in the reverse process we need to multiply.

This method has very limited uses. Generally we factor to find the roots and here we need the roots first in order to factor. A lot of text books say expressions like this can't be factored, but this is not the case, all quadratics can be factored. It should be expressed that some quadratics can only be factored if the roots are known first and not that they can't be factored.