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Question 186400This question is from textbook
: State the possible number of positive real zeros, negative real zeros & imaginary zero of this function
6. F(x)= 5x^3+8x^3-4x+3
This question is from textbook
Answer by kev82(151)   (Show Source):
You can put this solution on YOUR website! I will assume you mean 8x^2 instead of 8x^3
Well, the first thing to note is that it is a cubic with real coefficients. By the fundamental theorem of algebra it will have 3 zeroes, and because it has real coefficients, then imaginary zeroes have to come in a conjugate pair, so either zero or two of those. The easiest thing to do would be to draw it and see, but let's take a more mathematical approach. We can use the turning points along with the intermediate value theorem to figure out things.
The turning points are the solution to dF/dx = 0. dF/dx = 15x^2+16x-4. Solving this gives approximately 0.21 and -1.28. Now in the function these evaluate to 2.56 and 10.75 respectively.
So lets think about this, the leading term of the function is +x^3 so for sufficiently negative x the function is negative, The function keeps on steadily increasing until it hits the first turning point at -1.28. This has a y value of 10.75 so by the intermediate value theorem, to get from very negative to 10.75 we have a real root, somewhere to the left of -1.28. The function then has to decrease to get down to 2.56 which is the next turning point. Note it is impossible for us to have crossed the axis here because the function is continuous and zero is not between 10.75 and 2.56. After this turning point we increase again and the leading term takes over heading us to +infty. By the IVT there is no zero between 2.56 and +infty so there is only one real zero, and it is negative because it must be less than 1.28.
So in conclusion there is one negative real zero, no positive real zeroes, and two imaginary zeroes. making 3 zeroes in total.
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