How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

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How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

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Answer 1 · 2016-10-16T18:32:35

$f (x) = 3 x^{3} - 5 x^{2} - 47 x - 15$ $f(x)=3x^3-5x^2-47x-15$

Explanation:

If the zero is c, the factor is (x-c).

So for zeros of $- 3, - \frac{1}{3}, 5$ $-3,-1/3, 5$ , the factors are

$(x + 3) (x + \frac{1}{3}) (x - 5)$ $(x+3)(x+1/3)(x-5)$

Let's take a look at the factor $(x + \frac{1}{3})$ $(x+color(blue)1/color(red)3)$ . Using the factor in this form will not result in integer coefficients because $\frac{1}{3}$ $1/3$ is not an integer.

Move the $3$ $color(red)3$ in front of the x and leave the $1$ $color(blue)1$ in place: $(3 x + 1)$ $(color(red)3x+color(blue)1)$ .

When set equal to zero and solved, both
$(x + \frac{1}{3}) = 0$ $(x+1/3)=0$ and $(3 x + 1) = 0$ $(3x+1)=0$ result in $x = - \frac{1}{3}$ $x=-1/3$ .

$f (x) = (x + 3) (3 x + 1) (x - 5)$ $f(x)=(x+3)(3x+1)(x-5)$

Multiply the first two factors.

$f (x) = (3 x^{2} + 10 x + 3) (x - 5)$ $f(x)=(3x^2+10x+3)(x-5)$

Multiply/distribute again.

$f (x) = 3 x^{3} + 10 x^{2} + 3 x - 15 x^{2} - 50 x - 15$ $f(x)=3x^3+10x^2+3x-15x^2-50x-15$

Combine like terms.

$f (x) = 3 x^{3} - 5 x^{2} - 47 x - 15$ $f(x)=3x^3-5x^2-47x-15$

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How do you write a polynomial function of least degree with integral coefficients that has the given zeros -3, -1/3, 5?

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