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
May 19, 2017

Answer:

The "possible" rational zeros are:

#+-1/5, +-1, +-7/5, +-7#

The actual zeros are:

#1, 1, 1/5, 7#

Explanation:

Given:

#f(x) = 5x^4-46x^3+84x^2-50x+7#

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #7# and #q# a divisor of the coefficient #5# of the leading term.

That means that the only possible rational zeros are:

#+-1/5, +-1, +-7/5, +-7#

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Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients is #+ - + - +#. By Descartes' Rule of Signs, since this has #4# changes of sign, we can deduce that #f(x)# has #4#, #2# or #0# positive real zeros.

The pattern of the signs of the coefficients of #f(-x)# is #+ + + + +#. With no changes of sign, we can deduce that #f(x)# has no negative real roots.

So the only possible rational zeros are:

#1/5, 1, 7/5, 7#

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Sum of coefficients shortcut

Note that the sum of the coefficients of #f(x)# is zero. That is:

#5-46+84-50+7 = 0#

Hence we can deduce that #x=1# is a zero and #(x-1)# a factor:

#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 #x=1# is a zero again and #(x-1)# a factor:

#5x^3-41x^2+43x-7 = (x-1)(5x^2-36x+7)#

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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 #x=1/5# is a zero and #(5x-1)# a factor:

#5x^2-36x+7 = (5x-1)(x-7)#

So the last zero is #x=7#