the polynomial f(x) = x^3 + 2x - 11 has a real zero between which 2 consecutive integers: a: 0 and 1 b: 1 and 2 c: 2 and 3 d: 3 and 4 e: 4 and 5

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the polynomial f(x) = x^3 + 2x - 11 has a real zero between which 2 consecutive integers: a: 0 and 1 b: 1 and 2 c: 2 and 3 d: 3 and 4 e: 4 and 5

Mathematics
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b. 1 and 2 In order for there to be a real zero the graph most likely went from a y-value that is negative to a y-value that is positive. When you cross from negative to positive you must pass across the x-axis which is a zero. If you put in 1 for all x's - you get out -8 so the point (1,-8) is on the graph. If you put in 2 for all x's - you get out 1 so the point (2,1) is on the graph. If you draw a line between these two points, you will see it crossed the x-axis.
Use the quadratic formula to find your roots first:\[x=\frac{-2\pm \sqrt{48}}{2}=\sqrt{-2 \pm 4\sqrt{3}}{2}=-1 \pm 2\sqrt{3} \approx -4.464,2.464\]
can you use the quadratic here with the x^3?

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sorry - i misread - i'm off sick today...for a reason it seems
it's ok, hope you feel better!
thanks
the best way is then to just plug the values to find one positive and 1 negative?
blexthing is right
Just plug each pair of numbers into your equation and if you get two function values that change sign, it MUST be the case that the function cut the x-axis (i.e. a root exists in that interval).
ok, got it, thank you all!
the intermediate value theorem kicks retrice
welcome

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