# FINDING THE ROOTS OF POLYNOMIALS

After that I solved the equation \$x^3+10x^2+35x+50=0\$and found the integer solution \$-5\$and I divided the polynomial to \$x+5\$ & got the answer \$x^2+5x+10\$ and factored it like this:

\$x(x+5)(x^2+5x+10)\$

I want an easier way to solve it; which way would you recommend?  Since \$x=0\$ & \$x=-5\$ are roots of the given equation,\$\$ (x+1)(x+2)(x+3)(x+4)-24 = x(x+5)cdot q(x) ag1 \$\$where \$q(x)\$ is a monic second-degree polynomial. We may notice that, by De l"Hopital"s rule,\$\$ q(0) = lim_x o 0frac(x+1)(x+2)(x+3)(x+4)-24x(x+5)=frac24,H_45=10 ag2\$\$and if \$q(x)=x^2+Kx+10\$, in order that the coefficient of \$x^3\$ is the same in both sides of \$(1)\$\$\$ 1+2+3+4 = K+5 ag3 \$\$i.e. \$K=5\$, has to lớn hold.

Bạn đang xem: Finding the roots of polynomials Since the polynomial is symmetric around \$x+2.5\$ let us set \$y=x+2.5\$.

Then\$\$(x+1)(x+2)(x+3)(x+4)-24=(y-frac32)(y-frac12)(y+frac12)(y+frac32)-24\=(y^2-frac14)(y^2-frac94)-24=y^4-frac52y^2-frac37516=(y^2+frac154)(y^2-frac254)\$\$ \$(x+1)(x+2)(x+3)(x+4) - 24 =(x^2 +5x +4)(x^2+5x+6) -24=Y\$

Let \$x^2+5x=t\$.

So, \$Y=(t+4)(t+6) - 24 = t^2+10t=t(t+10)\$Implying, \$Y=(x^2+5x)(x^2+5x+10)=x(x+5)(x^2+5x+10)\$.

Xem thêm: Top 5 Cách Phối Áo Hoodie Với Chân Váy Không Bao Giờ Lỗi Mốt Thanks for contributing an answer to lớn ccevents.vnematics Stack Exchange!

But avoid

Asking for help, clarification, or responding lớn other answers.Making statements based on opinion; back them up with references or personal experience.

Use ccevents.vnJax to lớn format equations. ccevents.vnJax reference.

Xem thêm: Câu 33 Hỗn Hợp X Gồm Ancol Metylic Ancol Etylic Và Glixerol, Hỏi Đáp 24/7