Question

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given `f(x)=x^3+7x^2+7x-15`?

Answer 1

Use Descartes' Rule of Signs to find that `f(x)` has `1` positive Real zero and `0` or `2` negative zeros.

Further find all (rational) zeros: `x=1`, `x=-3` and `x=-5`

Explanation:

Given:

`f(x) = x^3+7x^2+7x-15`

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Descartes' Rule of Signs

The pattern of signs of the coefficients is `+ + + -`. With one change of signs, that means that this cubic has exactly `1` positive Real zero.

The pattern of signs of coefficients of `f(-x)` is `- + - -`. With two changes of sign, that menas that `f(x)` has `0` or `2` negative Real zeros.

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Rational Roots Theorem

Since `f(x)` is expressed in standard form and has integer coefficients we can apply the rational roots theorem to assert that 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 `-15` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-3, +-5, +-15`

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Sum of coefficients shortcut

The sum of the coefficients of `f(x)` is `0`. That is:

`1+7+7-15 = 0`

Hence `f(1) = 0` and `(x-1)` is a factor:

`x^3+7x^2+7x-15 = (x-1)(x^2+8x+15)`

Note that `3+5 = 8` and `3*5 = 15`, so we find:

`x^2+8x+15 = (x+3)(x+5)`

So the remaining two zeros are `x=-3` and `x=-5` as we might suspect.