Is #x-1# a factor of #P(x)=x^567-3x^400+x^9+2#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Cesareo R. Jan 16, 2017 #(x-1)# is not a factor of #P(x)# Explanation: If #x-1# were a factor of #P(x)# then #P(x)=(x-1)q(x)# and then #P(1)=(1-1)q(1)=0# but as can be verified #P(1)=1-3+1+2=1# so #(x-1)# is not a factor of #P(x)# Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 2676 views around the world You can reuse this answer Creative Commons License