How do you find the zeros of 4x4+3x3+2x2−3x+4?

1 Answer
George C.
Feb 17, 2016

See explanation...

Explanation:

This is an interesting question due to the form of the quartic.

First note the symmetry/anti-symmetry of the coefficients. This results in the property that if x=r is a zero then x=−1r is also a zero:

Let f(x)=4x4+3x3+2x2−3x+4

Then:

r4f(−1r)=r4(4r−4−3r−3+2r−2+3r−1+4)

=4−3r+2r2+3r3+4r4=f(r)

Taking such a pair of zeros, we find:

(x−r)(x+1r)=x2+(1r−r)x−1

So f(x) will have a quadratic factorization of the form:

4x4+3x3+2x2−3x+4

=4(x2+(1r1−r1)x−1)(x2+(1r2−r2)x−1)

=(2x2+2(1r1−r1)x−2)(2x2+2(1r2−r2)x−2)

Therefore consider:

4x4+3x3+2x2−3x+4

=(2x2+ax−2)(2x2+bx−2)

=4x4+2(a+b)x3+(ab−8)x2−2(a+b)x+4

Equating coefficients we find:

2(a+b)=3

ab=10

Hence (to cut a long story a little shorter) we can find:

a=34+√1514i

b=34−√1514i

So the zeros of f(x) are the roots of the two quadratic equations:

2x2+(34+√1514i)x−2=0

2x2+(34−√1514i)x−2=0

Using the quadratic formula to find the roots, we obtain formulae including the square root of a Complex number.

x=116(−3−√151i±√114+6√151i)

x=116(−3+√151i±√114−6√151i)

To convert these to a+bi form use the method described in http://socratic.org/questions/how-do-you-find-the-square-root-of-an-imaginary-number-of-the-form-a-bi

The square roots of c+di are:

±⎛⎝⎛⎝√√c2+d2+c2⎞⎠+⎛⎝d|d|√√c2+d2−c2⎞⎠i⎞⎠

Hence (using principal square root based on Arg(z)∈(−π,π]):

√114+6√151i=√48√2+57+(√48√2−57)i

√114−6√151i=√48√2+57−(√48√2−57)i

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