anonymous
  • anonymous
What are the possible number of positive real, negative real, and complex zeros of f(x) = -7x^4 - 12x^3 + 9x^2 - 17x + 3? i do not know where to start really on this problem
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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whpalmer4
  • whpalmer4
Descartes' Rule of Signs should give you the answer here, do you know it?
amoodarya
  • amoodarya
-7*x^4-12*x^3+9*x^2-17*x+3 = 0 => -7*(x+2.595)*(x-0.19)*(x^2-0.69*x+0.868) = 0 Real roots: Root 1: Root 2: Complex roots: Root 3: Root 4: -2.595 0.19 0.345+i*0.865 0.345-i*0.865
whpalmer4
  • whpalmer4
Here's how to use DRoS: write your polynomial with the terms in descending order by degree, just as you have it. To find the total number of roots, look at the exponent of the highest order term. That's the number of roots you'll have, though some of them may have the same value. Next, scan the polynomial from left to right, counting the number of times the sign of the coefficients changes. You've got - - + - +, so there are 3 sign changes. Now rewrite the polynomial, substituting (-x) wherever you see x in the original. Here we'll have \[-7(-x)^4 - 12(-x)^3 + 9(-x)^2 - 17(-x) + 3= -7x^4+12x^3+9x^2+17x+3\] All of the terms with odd exponents will have their coefficients change sign. Count the sign changes in the new polynomial, again scanning left to right. The first count you got is the maximum number of positive real roots. It may be reduced by a multiple of 2 to account for possible complex conjugate roots. The second count is the maximum number of negative real roots. Again, it may be reduced by a multiple of 2 to account for possible complex conjugate roots. The number of complex roots will be the difference between the number of roots, and the number of positive+negative real roots. It should always be a multiple of 2 if the coefficients of the polynomial are all real numbers. Counting the first polynomial, we have 3 sign changes, so we have either 3 positive real roots, or 1 positive real root. Counting the second polynomial, we have 1 sign change, so there is 1 negative real root. Altogether, we have as possibilities either: 3 positive real + 1 negative real roots 1 positive real, 1 negative real, and 2 complex roots By graphing, solving, or using calculus we can determine which of those possibilities is the one that matches this particular equation.

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anonymous
  • anonymous
thank you guys so much
anonymous
  • anonymous
for the explaination made alot of sense

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