# 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?

Dec 13, 2016

Use Descartes' Rule of Signs to find that $f \left(x\right)$ 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:

$f \left(x\right) = {x}^{3} + 7 {x}^{2} + 7 x - 15$

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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 $f \left(- x\right)$ is $- + - -$. With two changes of sign, that menas that $f \left(x\right)$ has $0$ or $2$ negative Real zeros.

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Rational Roots Theorem

Since $f \left(x\right)$ is expressed in standard form and has integer coefficients we can apply the rational roots theorem to assert that any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ 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:

$\pm 1 , \pm 3 , \pm 5 , \pm 15$

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Sum of coefficients shortcut

The sum of the coefficients of $f \left(x\right)$ is $0$. That is:

$1 + 7 + 7 - 15 = 0$

Hence $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

${x}^{3} + 7 {x}^{2} + 7 x - 15 = \left(x - 1\right) \left({x}^{2} + 8 x + 15\right)$

Note that $3 + 5 = 8$ and $3 \cdot 5 = 15$, so we find:

${x}^{2} + 8 x + 15 = \left(x + 3\right) \left(x + 5\right)$

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