# 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

#### 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