What are all the possible rational zeros for #f(x)=5x^4-46x^3+84x^2-50x+7# and how do you find all zeros?
1 Answer
Explanation:
Given:
#f(x) = 5x^4-46x^3+84x^2-50x+7#
Rational Roots Theorem
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/5, +-1, +-7/5, +-7#
Descartes' Rule of Signs
Note that the pattern of the signs of the coefficients is
The pattern of the signs of the coefficients of
So the only possible rational zeros are:
#1/5, 1, 7/5, 7#
Sum of coefficients shortcut
Note that the sum of the coefficients of
#5-46+84-50+7 = 0#
Hence we can deduce that
#5x^4-46x^3+84x^2-50x+7=(x-1)(5x^3-41x^2+43x-7)#
Looking at the coefficients of the remaining cubic, we find:
#5-41+43-7 = 0#
Hence
#5x^3-41x^2+43x-7 = (x-1)(5x^2-36x+7)#
Let's try the remaining rational possibilities with the remaining quadratic:
We find:
#5(color(blue)(1/5))^2-36(color(blue)(1/5))+7 = 5/25-36/5+7 = (1-36+35)/5 = 0#
So
#5x^2-36x+7 = (5x-1)(x-7)#
So the last zero is