Question

What are the roots of `(x+1)(2x+3)(2x+5)(x+3) = 945` ?

Answer 1

The roots are:

`x = 2`, `x = -6`, `x = -2+-sqrt(59)/2i`

Explanation:

Given:

`(x+1)(2x+3)(2x+5)(x+3) = 945`

Notice that:

`(x+1)` and `(x+3)` differ by `2`

`(2x+3)` and `(2x+5)` differ by `2`

What are the factors of `945`?

The prime factorisation is:

`945 = 3^3*5*7`

So we have:

`945 = 3*5*7*9 = (color(red)(2)+1) * (color(red)(2)+3) * (2*color(red)(2)+3) * (2*color(red)(2)+5)`

So one solution is `x=2`

What about other solutions?
Let `t = x+2`

Then:

`0 = (x+1)(2x+3)(2x+5)(x+3) - 945`

`color(white)(0) = (t-1)(2t-1)(2t+1)(t+1) - 945`

`color(white)(0) = (t^2-1)(4t^2-1) - 945`

`color(white)(0) = 4t^4-5t^2-944`

We know that `t = 2+2 = 4` is a root, so `t=-4` is also a zero `(t-4)(t+4) = t^2-16` is a factor:

`0 = 4t^4-5t^2-944 = (t^2-16)(4t^2+59)`

Hence the other two roots are `t = +-sqrt(59)/2i`

So the roots of our transformed quartic are:

`t = +-4` and `t = +-sqrt(59)/2i`

Then `x = t - 2` so the roots of the original equation are:

`x = 2`, `x = -6`, `x = -2+-sqrt(59)/2i`