Question

Find all roots of the equation x4-2x^3+14x^2-18x+45=0

Answer 1

We can write $14x^2$ as $5x^2+9x^2$ so:

$x^4-2x^3+14x^2-18x+45=0\\\\x^4-2x^3+5x^2+9x^2-18x+45=0\\\\x^2\big(x^2-2x+5\big)+9\big(x^2-2x+5\big)=0\\\\\big(x^2+9\big)\big(x^2-2x+5\big)=0$

$x^2+9=0\qquad\qquad\vee\qquad\qquad x^2-2x+5=0\\\\ x^2=-9\qquad\qquad\vee\qquad\qquad x^2-2x+1+4=0\\\\x^2=i^2\cdot3^2\qquad\qquad\vee\qquad\qquad (x-1)^2+4=0\\\\x^2=(3i)^2\quad|\sqrt{(\ldots)}\qquad\qquad\vee\qquad\qquad (x-1)^2=-4$

$x=3i\qquad\vee\qquad x=-3i\qquad\vee\qquad (x-1)^2=i^2\cdot2^2\\\\ x=3i\qquad\vee\qquad x=-3i\qquad\vee\qquad (x-1)^2=(2i)^2\quad|\sqrt{(\ldots)}\\\\ x=3i\quad\vee\quad x=-3i\quad\vee\quad x-1=2i\quad\vee\quad x-1=-2i\\\\ \boxed{x=3i\quad\vee\quad x=-3i\quad\vee\quad x=1+2i\quad\vee\quad x=1-2i}$

Answer 2

Quadratic factors of a quartic can be found with some trial and error. You can make use of the expansion ...

$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$

to write four equations in a, b, c, and d. These can be somewhat messy to solve, but some "guess and check" will get to a solution without too much trouble.

Here, the equations you get are ...

$a+c=-2\\ac+b+d=14\\ad+bc=-18\\bd=45$

Since 45 has only 6 divisors, the last of these gives rise to 6 possibilities to check: (b, d) = (±1, ±45), (±3, ±15), and (±5, ±9).

We choose to check (b, d) = (5, 9) first. Then our equations reduce to

$a+c=-2\\ac=0\\9a+5c=-18$

and it becomes pretty clear that c=0, a=-2 is a solution to these. Then our factorization is

$\left(x^2-2x+5\right)\left(x^2+9\right)=0$

The first quadratic can be written as

... (x-1)² +4 = 0

so has roots that can be found by subtracting 4, taking the square root, then adding 1.

... x = 1 ± √-4 = 1 ±2i

The second factor's roots can be found by subtracting 9 and taking the square root.

... x = √(-9) = ±3i

So the roots of the original quartic equation are ...

... {1 -2i, 1 +2i, -3i, +3i}