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