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bonufazy [111]
3 years ago
5

What are the possible numbers of positive, negative, and complex zeros of

Mathematics
1 answer:
o-na [289]3 years ago
3 0
Answer:

Look at changes of signs to find this has <span>1 </span> positive zero, <span>1 </span> or <span>3 </span> negative zeros and <span>0 </span> or <span>2 </span> non-Real Complex zeros.

Then do some sums...

Explanation:

<span><span><span>f<span>(x)</span></span>=−3<span>x4</span>−5<span>x3</span>−<span>x2</span>−8x+4</span> </span>

Since there is one change of sign, <span><span>f<span>(x)</span></span> </span> has one positive zero.

<span><span><span>f<span>(−x)</span></span>=−3<span>x4</span>+5<span>x3</span>−<span>x2</span>+8x+4</span> </span>

Since there are three changes of sign <span><span>f<span>(x)</span></span> </span> has between <span>1 </span> and <span>3 </span> negative zeros.

Since <span><span>f<span>(x)</span></span> </span> has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so <span><span>f<span>(x)</span></span> </span> has exactly <span>1 </span> or <span>3 </span> negative zeros counting multiplicity, and <span>0 </span> or <span>2 </span> non-Real Complex zeros.

<span><span>f'<span>(x)</span>=−12<span>x3</span>−15<span>x2</span>−2x−8</span> </span>

Newton's method can be used to find approximate solutions.

Pick an initial approximation <span><span>a0</span> </span>.

Iterate using the formula:

<span><span><span>a<span>i+1</span></span>=<span>ai</span>−<span><span>f<span>(<span>ai</span>)</span></span><span>f'<span>(<span>ai</span>)</span></span></span></span> </span>

Putting this into a spreadsheet and starting with <span><span><span>a0</span>=1</span> </span> and <span><span><span>a0</span>=−2</span> </span>, we find the following approximations within a few steps:

<span><span><span>x≈0.41998457522194</span> </span><span><span>x≈−2.19460208831628</span> </span></span>

We can then divide <span><span>f<span>(x)</span></span> </span> by <span><span>(x−0.42)</span> </span> and <span><span>(x+2.195)</span> </span> to get an approximate quadratic <span><span>−3<span>x2</span>+0.325x−4.343</span> </span> as follows:

Notice the remainder <span>0.013 </span> of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

<span><span><span>−3<span>x2</span>+0.325x−4.343</span> </span><span><span><span>Δ=<span>b2</span>−4ac=<span>0.3252</span>−<span>(4⋅−3⋅−4.343)</span>=0.105625−52.116=</span><span>−52.010375</span></span> </span></span>

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly <span>2 </span> non-Real Complex zeros, <span>1 </span> positive zero and <span>1 </span> negative one.

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