Numeric Approximation of a Zero: The Quicker Way

From the graph, I can see that the zero is definitely between x = 1 and x = 2.

I'll find the corresponding y-values:

This confirms what the graph told me, but this also tells me that the zero is likely to be a lot closer to x = 1 than to x = 2. Why? Because −3 is a lot closer to zero than is 32.

Since the zero is closer to x = 1, I'll try x = 1.4.

This gets me a lot closer to the axis, but this y-value is positive so the zero is still lower. I'll try x = 1.2:

That's nearly as negative as the last y-value was positive. Okay, so the zero is above x = 1.2 and below x = 1.4. I'll try x = 1.3:

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Oooo; that's pretty good. Okay, so I'm really close to the polynomial's zero, but I'm still a little high. I'll edge down by hundredths and try x = 1.29:

That took me further from the axis, and below. So the zero is higher, and probably a lot closer to 1.300 than to 1.290. So I'll try x = 1.297:

This y-value is closer to zero, but I'm still too low, and x = 1.300 got me a bit closer still, so I'll try x = 1.299:

I'm getting closer, but I'm still too low. I've noticed that 1.3000 was not quite as close to the zero as was 1.2990, but the polynomial also seems to grow faster after the zero than before it, so I'll split the difference and see what I get:

Okay, now I'm too high again, but I'm getting pretty close. Lemme edge down a bit:

Finally! I've managed to bracket the zero between two x-values that have the same first three decimal places. That's all the accuracy I need, so I'm done.

The first thing I have to do is figure out the general area of the zero. Without a graph and without interval endpoints being given to me, I'll have to use what I know about polynomials to judge the values I get. I'll try a few values in the vicinity of the origin:

Bingo! I've found a sign change, so a zero of the polynomial is somewhere between x = −3 and x = −2. Also, the polynomial is above the axis at the lower x-value, so I know the graph goes down as it crosses the axis. Then a negative y-value will mean that my x-value was too high, and a positive y-value will mean that my x-value was too low. In other words, the signs on the outputs will tell me in which direction I need to move.

Since y = −2.5 (at x = −2) is a lot closer to zero than is y = 36 (at x = −3), I'm going to guess that the zero must be a lot closer to x = −2, so I'll test some values in the tenths:

Okay, now I know the zero is between x = −2.2 and x = −2.1, and it looks like it's closer to x = −2.1. I'll work my way down into the hundredths:

The sign change tells me that the zero is bracketed by x = −2.14 and x = −2.13, with the polynomial being closer to the axis at x = −2.13. I'll test values in the thousandths:

Now I have the zero bracketed between −2.135 and −2.134, numbers that have the same first two decimal places, so I'm done.