How do you write a polynomial function of least degree with integral coefficients that has the given zeros 3, 2, -1?

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How do you write a polynomial function of least degree with integral coefficients that has the given zeros 3, 2, -1?

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Answer 1 · 2016-07-18T05:32:02

$y = (x - 3) (x - 2) (x + 1)$ $y=(x-3)(x-2)(x+1)$
Also
$y = x^{3} - 4 x^{2} + x + 6$ $y=x^3-4x^2+x+6$

Explanation:

From the given zeros 3, 2, -1

We set up equations $x = 3$ $x=3$ and $x = 2$ $x=2$ and $x = - 1$ $x=-1$ . Use all these as factors equal to the variable y.

Let the factors be $x - 3 = 0$ $x-3=0$ and $x - 2 = 0$ $x-2=0$ and $x + 1 = 0$ $x+1=0$

$y = (x - 3) (x - 2) (x + 1)$ $y=(x-3)(x-2)(x+1)$

Expanding

$y = (x^{2} - 5 x + 6) (x + 1)$ $y=(x^2-5x+6)(x+1)$
$y = (x^{3} - 5 x^{2} + 6 x + x^{2} - 5 x + 6)$ $y=(x^3-5x^2+6x+x^2-5x+6)$
$y = x^{3} - 4 x^{2} + x + 6$ $y=x^3-4x^2+x+6$

Kindly see the graph of $y = x^{3} - 4 x^{2} + x + 6$ $y=x^3-4x^2+x+6$ with zeros at $x = 3$ $x=3$ and $x = 2$ $x=2$ and $x = - 1$ $x=-1$

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How do you write a polynomial function of least degree with integral coefficients that has the given zeros 3, 2, -1?

1 Answer

Explanation:

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