Question

How do you factor `y^4+8y^3+12`?

Answer 1

If the question is correct as stands then the answer is complicated, but here's a sketch of how you can do it...

Explanation:

`y^4+8y^3+12`

Let `t = y+2`

Then:

`y^4+8y^3+12 = (y+2)^4-24(y+2)^2+64(y+2)-36`

`color(white)(y^4+8y^3+12) = t^4-24t^2+64t-36`

`color(white)(y^4+8y^3+12) = (t^2-at+b)(t^2+at+c)`

`color(white)(y^4+8y^3+12) = t^4+(b+c-a^2)t^2+a(b-c)t+bc`

Equating coefficients:

`{ (b+c = a^2-24), (b-c = 64/a), (bc=-36) :}`

Hence:

`a^4-48a^2+576 = (a^2-24)^2 = (b+c)^2 = (b-c)^2+4bc = 4096/a^2-144`

Hence:

`0 = (a^2)^3-48(a^2)^2+720(a^2)-4096`

`color(white)(0) = (a^2-16)^3-48(a^2-16)-768`

Using Cardano's method, let `a^2-16 = u+v`

Then:

`u^3+v^3+3(uv-16)(u+v)-768 = 0`

Add the constraint `v = 16/u` to eliminate the term in `(u+v)` and get:

`(u^3)^2-768(u^3)+4096 = 0`

Using the quadratic formula:

`u^3 = (768+-sqrt(768^2-4(1)(4096)))/2`

`color(white)(u^3) = 384+-64sqrt(35)`

`color(white)(u^3) = 4^3(6+-sqrt(35))`

Hence:

`a^2 = 16+4root(3)(6+sqrt(35))+4root(3)(6-sqrt(35))`

We can use the positive square root:

`a = 2sqrt(4+root(3)(6+sqrt(35))+root(3)(6-sqrt(35)))`

Then:

`b = 1/2(a^2-24+64/a)`

`c = 1/2(a^2-24-64/a)`

Leaving us with two quadratics to solve:

`t^2-at+b = 0`

`t^2+at+c = 0`

yielding `4` roots: `t_1, t_2, t_3, t_4`.

Then the zeros of our original quartic are given by:

`y_n = t_n-2`

Hence we find a factorisation:

`y^4+8y^3+12 = (y-y_1)(y-y_2)(y-y_3)(y-y_4)`

where `y_1, y_2, y_3, y_4` are the four zeros.