53.2k views
4 votes
What are the real roots of 2x^3+9x^2+4x-15=0

1 Answer

4 votes
Well, luckily it is apparent that (x-1) is a root because when x=1 the equation is equal to zero. So we can divide the equation by that factor to find the other roots.

(2x^3+9x^2+4x-15)/(x-1)
2x^2 r 11x^2+4x-15
11x r 15x-15
15 r 0

(x-1)(2x^2+11x+15)=0

(x-1)(2x^2+6x+5x+15)=0

(x-1)(2x(x+3)+5(x+3))=0

(x-1)(2x+5)(x+3)=0

So the roots are x= -3, -2.5, 1
User Chris Marie
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories