I was reading a wiki page on brilliant.org titled Factoring by Substitution. I have a few questions regarding one of the solved examples given there
The question says: Solve for all the roots of the following polynomial $$x^4-5ix^3+19x^2-125ix-150$$ Let the above polynomial be represented by $p(x)$. After making the substitution $a=ix$ and some grouping, the following result is stated $$p(x)=(a^3-19a+30)(a-5)$$ Then the author states: "Using some logic, since there's a gap in powers ($a^3$ followed by $a$), there should be a difference of squares. By remainder theorem, since 5 is a solution, we can check -5."
Firstly, I don't understand why a gap in powers implies the existence of a difference in squares. Could anyone please explain this with some examples?
Secondly, how does the remainder theorem come into picture here? Even if it is applicable..the gap in powers is present in the first term, and 5 is not a factor of the first expression in the product..
On the whole... I don't understand the entire statement that is made which I've included in double quotations(" ") above...Could anyone please help me with this?
Link to the wiki page: https://brilliant.org/wiki/factor-polynomials-by-substitution/?subtopic=polynomials&chapter=polynomial-factoring
Thanks for any answers!!
For the polynomial $a^3-19a+30$ it means the following reasoning.
Let $t$ be a root of the polynomial.
Thus, there is a factor $a-t$, but since the coefficient before $a^2$ is equal to $0$, we need to get a difference of squares $a^2-t^2$: $$a^3-19a+30=a^3-ta^2+ta^2-t^2a+t^2a-19a+30=$$ $$=a^2(a-t)+ta(a-t)+t^2a-19a+30=$$ $$=a(a-t)(a+t)+t^2a-19a+30=a(a^2-t^2)+t^2a-19a+30.$$
For $x^4-5ix^3+19x^2-125ix-150$ by your hint we obtain:
let $x=ai$.
Thus, $$x^4-5ix^3+19x^2-125ix-150=a^4-5a^3-19a^2+125a-150.$$ Now, easy to see that $5$ and $-5$ are roots.
Thus, we have a factor $a^2-25$ and $$a^4-5a^3-19a^2+125a-150=a^4-25a^2-5a^3+125a+6a^2-150=$$ $$=(a^2-25)(a^2-5a+6)=(a-5)(a+5)(a-2)(a-3)=$$ $$=(ai-2i)(ai-3i)(ai+5i)(ai-5i)=$$ $$=(x-2i)(x-3i)(x+5i)(x-5i).$$