How do you find all the zeros of h(x) = x^4 + 10x^3 + 26x^2 + 10x + 25h(x)=x4+10x3+26x2+10x+25?

1 Answer
George C.
Mar 20, 2016

Use the rational root theorem to help find the first zero, then factor by it and by grouping to find the other roots.

Explanation:

h(x) = x^4+10x^3+26x^2+10x+25h(x)=x4+10x3+26x2+10x+25

By the rational root theorem, any rational roots of h(x)=0h(x)=0 must be expressible in the form p/qpq for integers pp and qq where pp is a factor of the constant term 2525 and qq a factor of the coefficient 11 of the leading term.

That means that the only possible rational zeros are:

+-1±1, +-5±5, +-25±25

In addition, note that all of the coefficients of h(x)h(x) are positive. So it can have no positive zeros. So that just leaves the possibilities:

-1−1, -5−5, -25−25

We find:

h(-1) = 1-10+26-10+25 = 32h(−1)=1−10+26−10+25=32

h(-5) = 625-1250+650-50+25 = 0h(−5)=625−1250+650−50+25=0

So x = -5x=−5 is a zero and (x+5)(x+5) a factor:

x^4+10x^3+26x^2+10x+25x4+10x3+26x2+10x+25

=(x+5)(x^3+5x^2+x+5)=(x+5)(x3+5x2+x+5)

=(x+5)(x^2(x+5)+1(x+5))=(x+5)(x2(x+5)+1(x+5))

=(x+5)^2(x^2+1)=(x+5)2(x2+1)

So x=-5x=−5 is a double zero and the other zeros are x=+-ix=±i

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