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

##### 2 Answers

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

#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

Next note that since it contains no term in

#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

If there is a positive real root then choose that one and choose

Having found a usable value of

#{ (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

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.