4 years ago

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

1 answer:

8 0

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

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