Question

How do you solve `x^4+3x^3-28x^2+13x+42 = 0` ?

Answer 1

Here's a way forward (but it's not very nice)...

Explanation:

If you are faced with a quartic equation that has no rational roots and includes terms of odd degree, then you can solve it algebraically, but it typically gets rather messy.

Here's a generic method of solution...

Starting with:

`ax^4+bx^3+cx^2+dx+e = 0`

We can divide through by `a` to get:

`0 = x^4+b/ax^3+c/ax^2+d/ax+e/a`

`color(white)(0) = (x+b/(4a))^4+(8ac-3b^2)/(8a^2)(x+b/(4a))^2+(8a^2d+b^3-4abc)/(8a^3)(x+b/(4a))+(256a^3e-64a^2bd+16ab^2c-3b^4)/(256a^4)`

`color(white)(0) = t^4+a_1t^2+b_1t+c_1`

where:

`{ (t = x+b/(4a)), (a_1 = (8ac-3b^2)/(8a^2)), (b_1 = (8a^2d+b^3-4abc)/(8a^3)), (c_1 = (256a^3e-64a^2bd+16ab^2c-3b^4)/(256a^4)) :}`

If at this stage we find that `b_1 = 0`, then this can be tackled as a quadratic in `t^2`. From this point on assume the harder case where `b_1 != 0` ...

Next note that since it contains no term in `t^3`, whenever it is factored into a pair of monic quadratic factors, it must take the form:

`t^4+a_1t^2+b_1t+c_1 = (t^2-pt+q)(t^2+pt+r)`

`color(white)(t^4+a_1t^2+b_1t+c_1) = t^4+(q+r-p^2)t^2+p(q-r)t+qr`

So equating coefficients, we find:

`{ (q+r = p^2+a_1), (q-r = b_1/p), (qr=c_1) :}`

So we find:

`(p^2+a_1)^2 = (q+r)^2 = (q-r)^2+4qr = (b_1/p)^2+4c_1`

Hence:

`(p^2)^3+2a_1(p^2)^2+(a_1^2-4c_1)(p^2)-b_1^2 = 0`

Solving this as a cubic in `p^2` will give three roots, of which at least one will be real.

If there is a positive real root then choose that one and choose `p` as its positive square root. Otherwise choose the negative real root and `p` will be a pure imaginary square root.

Having found a usable value of `p`, we can deduce `q` and `r` from the previous equations:

`{ (q = 1/2((q+r)+(q-r)) = 1/2(p^2+a_1+b_1/p)), (r = 1/2((q+r)-(q-r)) = 1/2(p^2+a_1-b_1/p)) :}`

That gives us two quadratic equations to solve:

`t^2-pt+1/2(p^2+a_1+b_1/p) = 0`

`t^2+pt+1/2(p^2+a_1-b_1/p) = 0`

For each of the resulting four values of `t`, we can deduce the values of our original `x` by subtracting `b/(4a)`.

Answer 2

Use Newton Raphson method or such numerical techniques

Explanation:

graph{x^4+3x^3-28x^2+13x+42 [-10, 10, -10 10]}

As you can see all the roots are not whole numbers. Hence start with different initial guess and run a NR algorithm to get a solution.