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!

Please be sure to lớn answer the question. Provide details & share your research!

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

To learn more, see our tips on writing great answers.

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy và cookie policy

Site thiết kế / logo sản phẩm © 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Rev2022.12.16.43123

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device & disclose information in accordance with our Cookie Policy.