How do you find all the zeros of h(x) = x^4 + 10x^3 + 26x^2 + 10x + 25?
1 Answer
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+25
By the rational root theorem, any rational roots of
That means that the only possible rational zeros are:
+-1 ,+-5 ,+-25
In addition, note that all of the coefficients of
-1 ,-5 ,-25
We find:
h(-1) = 1-10+26-10+25 = 32
h(-5) = 625-1250+650-50+25 = 0
So
x^4+10x^3+26x^2+10x+25
=(x+5)(x^3+5x^2+x+5)
=(x+5)(x^2(x+5)+1(x+5))
=(x+5)^2(x^2+1)
So