# What are all the possible rational zeros for y=x^3+x^2-9x+7 and how do you find all zeros?

Sep 25, 2016

$x = 1$ or $x = - 1 \pm 2 \sqrt{2}$

#### Explanation:

$f \left(x\right) = {x}^{3} + {x}^{2} - 9 x + 7$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $7$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 7$

Since the sum of the coefficients of $f \left(x\right)$ is zero (i.e. $1 + 1 - 9 + 7 = 0$) then $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} + {x}^{2} - 9 x + 7 = \left(x - 1\right) \left({x}^{2} + 2 x - 7\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

as follows:

${x}^{2} + 2 x - 7 = {x}^{2} + 2 x + 1 - 8$

$\textcolor{w h i t e}{{x}^{2} + 2 x - 7} = {\left(x + 1\right)}^{2} - {\left(2 \sqrt{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 2 x - 7} = \left(\left(x + 1\right) - 2 \sqrt{2}\right) \left(\left(x + 1\right) + 2 \sqrt{2}\right)$

$\textcolor{w h i t e}{{x}^{2} + 2 x - 7} = \left(x + 1 - 2 \sqrt{2}\right) \left(x + 1 + 2 \sqrt{2}\right)$

Hence the remaining zeros are:

$x = - 1 \pm 2 \sqrt{2}$