How do you find all the zeros of f(x)=x4−9x3+24x2−6x−40?

1 Answer
George C.
May 8, 2016

x=−1, x=4, x=3±i

Explanation:

First notice that if you reverse the signs of the coefficients on the terms of odd degree then the sum is 0. That is:

1+9+24+6−40=0

So f(−1)=0 and (x+1) is a factor:

x4−9x3+24x2−6x−40=(x+1)(x3−10x2+34x−40)

Let g(x)=x3−10x2+34x−40

By the rational root theorem, any rational zeros of g(x) must be expressible in the form pq for integers p, q with p a divisor of the constant term −40 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40

In addition, note the g(−x)=−x3−10x2−34x−40 has all negative coefficients. So there are no negative zeros.

So the only possible rational zeros of g(x) are:

1,2,4,5,8,10,20,40

We find:

g(4)=64−160+136−40=0

So x=4 is a zero and (x−4) a factor:

x3−10x2+34x−40

=(x−4)(x2−6x+10)

The remaining quadratic factor has negative discriminant, so its zeros are Complex, but we can factor it as a difference of squares:

x2−6x+10

=(x−3)2−9+10

=(x−3)2+1

=(x−3)2−i2

=((x−3)−i)((x−3)+i)

=(x−3−i)(x−3+i)

So the remaining zeros are:

x=3±i

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