How do you find all the zeros of f(x) = x^4 - 9x^3 + 24x^2 - 6x - 40f(x)=x4−9x3+24x2−6x−40?

1 Answer
George C.
May 8, 2016

x = -1x=−1, x = 4x=4, x = 3+-ix=3±i

Explanation:

First notice that if you reverse the signs of the coefficients on the terms of odd degree then the sum is 00. That is:

1+9+24+6-40 = 01+9+24+6−40=0

So f(-1) = 0f(−1)=0 and (x+1)(x+1) is a factor:

x^4-9x^3+24x^2-6x-40 = (x+1)(x^3-10x^2+34x-40)x4−9x3+24x2−6x−40=(x+1)(x3−10x2+34x−40)

Let g(x) = x^3-10x^2+34x-40g(x)=x3−10x2+34x−40

By the rational root theorem, any rational zeros of g(x)g(x) must be expressible in the form p/qpq for integers pp, qq with pp a divisor of the constant term -40−40 and qq a divisor of the coefficient 11 of the leading term.

That means that the only possible rational zeros are:

+-1±1, +-2±2, +-4±4, +-5±5, +-8±8, +-10±10, +-20±20, +-40±40

In addition, note the g(-x) = -x^3-10x^2-34x-40g(−x)=−x3−10x2−34x−40 has all negative coefficients. So there are no negative zeros.

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

1, 2, 4, 5, 8, 10, 20, 401,2,4,5,8,10,20,40

We find:

g(4) = 64-160+136-40 = 0g(4)=64−160+136−40=0

So x=4x=4 is a zero and (x-4)(x−4) a factor:

x^3-10x^2+34x-40x3−10x2+34x−40

=(x-4)(x^2-6x+10)=(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:

x^2-6x+10x2−6x+10

=(x-3)^2-9+10=(x−3)2−9+10

=(x-3)^2+1=(x−3)2+1

=(x-3)^2-i^2=(x−3)2−i2

=((x-3)-i)((x-3)+i)=((x−3)−i)((x−3)+i)

=(x-3-i)(x-3+i)=(x−3−i)(x−3+i)

So the remaining zeros are:

x = 3+-ix=3±i

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