What are the real roots of 2x^3+9x^2+4x-15=0
1 answer:
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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To your question, "K<-5", that is the answer. So the answer is K<-5
Yes the correct answer is D. Hope this helps :)
Answer:
-5/7
Step-by-step explanation:
Look at the graph. As we move from left to right from (-3, 2) to (4, -3), x increases by 7 and y decreases by 5. Thus, the slope is
m = rise/run = -5/7 .