# Doubts regarding tricks to factor polynomials

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?

• Check please coefficients of the polynomial. Maybe it means $x^4-5x^3+19x^2-125x-150$ or $x^4-5ix^3+19x^2-125ix-150$? Commented Apr 16, 2020 at 18:22
• Thanks for pointing out! I missed the $i$ in the 4th term! I have edited my question Commented Apr 16, 2020 at 18:26
• Your polynomial directly factors as $(x^2 - 6 - 5ix)(x^2 + 25)$. This seems more direct, without any tricks. Of course $x^2+25=(x+5i)(x-5i)$ and so on. Commented Apr 16, 2020 at 18:42
• I just computed $(x^2+ax+b)(x^2+cx+d)$ and compared coefficients. Very easy. No tricks. Commented Apr 16, 2020 at 18:44
• Perhaps "gaps of powers" here just refers to even or odd exponent of $x$, which plays a role if you substitute $a=xi$. Commented Apr 16, 2020 at 18:47

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 $$a^4-5a^3-19a^2+125a-150=a^4-25a^2-5a^3+125a+6a^2-150=$$ $$=(a^2-25)(a^2-5a+6)=(a-5)(a+5)(a-2)(a-3)=$$ $$=(ai-2i)(ai-3i)(ai+5i)(ai-5i)=$$ $$=(x-2i)(x-3i)(x+5i)(x-5i).$$

• Do you know anything about the gap in powers which was mentioned? Commented Apr 16, 2020 at 18:47
• @binarybitarray I added something. See now. Commented Apr 16, 2020 at 19:04