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descartes rule of sign for this one
can you help me find the answer?
\[f(x)=-2x^3-5x^2+6x+4\] coefficients are \(-2,-5,6,4\)
there is one "change of sign" for \(f(x)\) from \(-5\) to \(+6\)
ok what is next?
so there is one positive zero you count down by 2s and there cannot be -1 positive zero
oh btw there are 3 zeros all together because the degree is 3, although some may be complex or repeated
now we compute \(f(-x)\)
the sign of the coefficients of the terms of odd degree change, and the even degree stays the same the coefficients are \(2,-5,-6,4\)
now there are two "changes in sign" from 2 to -5 and from -6 to 4 is that more or less clear?
ok is that the final answer?
can you help me with one more? it is easier this time
oh no two changes in sign of \(f(-x)\) means either 2 or 0 negative zeros
oh ok sorry..please finish the problem
so final answer is this: 1 positive zero for sure either 2 negative zeros and therefore no complex zzeros or no negative zeros and two complex zeros
ok thanks..can you do one more? please?
this one is easier \[\sum_3^53n+2\]
put \(n=3\) get \(3\times 3+2=11\) then put \(n=4\) get \(3\times 4+2=14\) finally put \(n=5\) get \(17\) the "sigma" \(\sum\) means add them up
\[11+14+17\]is what you need to compute, and you are done