# How do you find the cubic polynomial function with two of its zeros 2 and -3+√2 and a y-intercept of 7?

May 31, 2016

$f \left(x\right) = - \frac{1}{2} {x}^{3} - 2 {x}^{2} + \frac{5}{2} x + 7$

#### Explanation:

If the cubic has rational coefficients, then its other zero must be the conjugate $- 3 - \sqrt{2}$ of $- 3 + \sqrt{2}$ and it will take the form:

$f \left(x\right) = a \left(x - 2\right) \left(x + 3 - \sqrt{2}\right) \left(x + 3 + \sqrt{2}\right)$

$= a \left(x - 2\right) \left({\left(x + 3\right)}^{2} - 2\right)$

$= a \left(x - 2\right) \left({x}^{2} + 6 x + 7\right)$

$= a \left({x}^{3} + 4 {x}^{2} - 5 x - 14\right)$

$= a {x}^{3} + 4 a {x}^{2} - 5 a x - 14 a$

In order that the $y$ intercept be $7$, we must have:

$f \left(0\right) = - 14 a = 7$

So $a = - \frac{1}{2}$ and our cubic function is:

$f \left(x\right) = - \frac{1}{2} {x}^{3} - 2 {x}^{2} + \frac{5}{2} x + 7$

graph{-1/2x^3-2x^2+5/2x+7 [-20.34, 19.66, -9.08, 10.92]}

Note that if we do not require $f \left(x\right)$ to have rational coefficients, then there is a whole family of other solutions.