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# How do you write a polynomial with zeros -6, 3, 5 and leading coefficient 1?

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#### Explanation

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#### Explanation:

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1
Jan 21, 2018

${x}^{3} - 2 {x}^{2} - 33 x + 90$

#### Explanation:

This has $3$ roots, so it will be a polynomial of degree $3$. If we factor a 3rd degree polynomial we get:

$\left(x + a\right) \left(x + b\right) \left(x + c\right)$ where a , b and c are constants:

We can find these constants using the known roots. If the roots are:

$\alpha = - 6$

$\beta = 3$

$\gamma = 5$

Since the constant term in each factor will have an opposite sign to the root, we have:

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$

Substituting $\alpha , \beta \mathmr{and} \gamma$ for root values we have:

$\left(x - \left(- 6\right)\right) \left(x - \left(3\right)\right) \left(x - \left(5\right)\right) = \left(x + 6\right) \left(x - 3\right) \left(x - 5\right)$

The coefficient of the first term is the product of the coefficients of the x terms in each factor, since these are all $1$, the coefficient of the first term will be $1$

$\left(x + 6\right) \left(x - 3\right) \left(x - 5\right) = {x}^{3} - 2 {x}^{2} - 33 x + 90$

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