بر اساس شرط داریم:
$$c={{abd-d^2}\over {b^2}}$$
سپس جایگذاری می کنیم:
$$x^5+ax^3+bx^2+cx+d
\\ =x^5+ax^3+bx^2+\frac{d(ab-d)}{b^2}x+d
\\
= x^5+ax^3+bx^2+\left( \frac{ad}{b}-\frac{d^2}{b^2} \right)x+d
\\
= x^5-\frac{d^2x}{b^2}+ax\left( x^2+\frac{d}{b} \right)+(bx^2+d)
\\
= x\left( x^4-\frac{d^2}{b^2} \right)+ax\left( x^2+\frac{d}{b} \right)+
(bx^2+d)
\\
= x\left( x^2+\frac{d}{b} \right) \left( x^2-\frac{d}{b} \right)+
ax\left( x^2+\frac{d}{b} \right)+b\left( x^2+\frac{d}{b} \right)
\\
= \left( x^2+\frac{d}{b} \right)
\left[ x\left( x^2-\frac{d}{b} \right)+ax+b \right]
\\
= \left( x^2+\frac{d}{b} \right)
\left[ x^3+\left( a-\frac{d}{b} \right)x+b \right]$$
