Question

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?

Answer 1

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`