Question

How do you factor `4x^4+81` ?

Answer 1

You can use Real or Complex arithmetic methods to find:

`4x^4+81 = (2x^2+6x+9)(2x^2-6x+9)`

Explanation:

Method 1 - Real arithmetic

First note that `4x^4+81` has no linear factors with Real coefficients, since `4x^4+81 >= 81 > 0` for all `x in RR`.

How about quadratic factors?

Since there is only a term in `x^4` and a constant term with positive sign, try the following factorisation:

`4x^4+81 = (2x^2+ax+b)(2x^2-ax+b)`

where `b = +-9`

Looking at the coefficient of the `x^2` term, we find:

`4b-a^2 = 0`

So if `a in RR`, then `b >= 0`, so `b = 9`

Then we find `a = +-6`

Hence:

`4x^4+81 = (2x^2+6x+9)(2x^2-6x+9)`

Method 2 - Complex arithmetic

Solve `4x^4+81 = 0` as follows:

Subtract `81` from both sides to get:

`4x^4 = -81`

Divide both sides by `4` to get:

`x^4 = -81/4`

Take square roots of both sides to get:

`x^2 = +-9/2i`

Now `+-sqrt(i) = +-(1 + i)/sqrt(2)`

So `+-sqrt(9/2i) = +-sqrt(9/2)(1+i)/sqrt(2) = +-(3+3i)/2`

and `+-sqrt(-9/2i) = +-isqrt(9/2i) = +-(i(3+3i))/2 = +-(3i-3)/2`

So

`4x^4+81`

`= 4(x+(3+3i)/2)(x+(3-3i)/2)(x-(3+3i)/2)(x-(3-3i)/2)`

`= (2(x+3/2+3/2i)(x+3/2-3/2i))(2(x-3/2-3/2i)(x-3/2+3/2i))`

`= (2x^2+6x+9)(2x^2-6x+9)`

Answer 2

`4x^4+81=0` has not real roots so necessarily has the structure

`4x^4+81=4(x^2+a_1x+b_1)(x^2+a_2x+b_2)`

(two pairs of complex conjugate roots)

Equating for all `x` we have the conditions

`{(81 - 4 b_1 b_2=0), (a_2 b_1 + a_1 b_2=0), (a_1 a_2 + b_1 + b_2=0), (a_1 + a_2=0):}`

with solution

`(a_1 = -3, b_1 = 9/2, a_2 = 3, b_2 = 9/2)` or

`4x^4+81 = 4(9/2 - 3 x + x^2) (9/2 + 3 x + x^2)`