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.
