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

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

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Answer 1 · 2016-05-31T22:16:35

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

Explanation:

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

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

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

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

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

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

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

$f (0) = - 14 a = 7$ $f(0) = -14a=7$

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

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

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

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

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

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