# How do you write a polynomial in standard form given zeros 3,-2,1/2?

Sep 26, 2016

$y = {x}^{3} - \frac{3}{2} {x}^{2} - \frac{11}{2} x + 3$

#### Explanation:

Given that the zeros of a polynomial are x = a , x = b and x = c.

Then $\left(x - a\right) , \left(x - b\right) \text{ and " (x-c)" are the factors}$

and $y = \left(x - a\right) \left(x - b\right) \left(x - c\right) \text{ is the polynomial}$

Here the zeros are $x = 3 , x = - 2 \text{ and } x = \frac{1}{2}$

$\Rightarrow \left(x - 3\right) , \left(x + 2\right) \text{ and " (x-1/2)" are the factors}$

$\Rightarrow y = \left(x - 3\right) \left(x + 2\right) \left(x - \frac{1}{2}\right) \text{ gives the polynomial}$

distribute the first 'pair' of brackets.

$= \left({x}^{2} - x - 6\right) \left(x - \frac{1}{2}\right)$

distributing and collecting like terms.

$= {x}^{3} - \frac{1}{2} {x}^{2} - {x}^{2} + \frac{1}{2} x - 6 x + 3$

$\Rightarrow y = {x}^{3} - \frac{3}{2} {x}^{2} - \frac{11}{2} x + 3 \text{ is the polynomial}$

Sep 26, 2016

The desired polynomial is $2 {x}^{3} - {x}^{2} - 13 x - 6$

#### Explanation:

A polynomial with zeros $p$, $q$ and $r$ can be written as $a \left(x - p\right) \left(x - q\right) \left(x - r\right)$

Hence a polynomial with zeros $3$, $- 2$ and $- \frac{1}{2}$ can be written as (here we have multiplied by $2$ to take care of $- \frac{1}{2}$ as zero and ensure all coefficients are integers).

$2 \left(x - 3\right) \left(x - \left(- 2\right)\right) \left(x - \left(- \frac{1}{2}\right)\right)$

= 2(x-3)(x+2))(x+1/2)

= $\left(x - 3\right) \left(x + 2\right) \left(2 x + 1\right)$

= $\left(x - 3\right) \left(x \left(2 x + 1\right) + 2 \left(2 x + 1\right)\right)$

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

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

= $2 {x}^{3} + 5 {x}^{2} + 2 x - 6 {x}^{2} - 15 x - 6$

= $2 {x}^{3} - {x}^{2} - 13 x - 6$

Sep 26, 2016

$\text{The Reqd. Poly. } = 2 a \left(2 {x}^{3} - 3 {x}^{2} - 11 x + 6\right) , a \ne 0 , a \in \mathbb{R} .$

#### Explanation:

We are given $3$ real zeroes of a Poly., say $p \left(x\right)$.

Hence, $p \left(x\right)$ has to be a cubic poly.

So, let, $p \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d , a \ne 0 ,$

To find the reqd. Poly., we can use Vieta's Rule, which relates the

zeroes of the Poly. with its co-effs.

Vieta's Rule for Cubic Poly. : If, $\alpha , \beta , \gamma$ are the

zeroes of a Cubic Poly. $p \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d , a \ne 0 ,$ then,

$\alpha + \beta + \gamma = - \frac{b}{a} \ldots \ldots \ldots . \left(1\right)$,

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \ldots \ldots \ldots . . \left(2\right)$,

$\alpha \beta \gamma = - \frac{d}{a} \ldots \ldots \ldots . \left(3\right)$

In our case, we have, by $\left(1\right) - \left(3\right)$,

$- \frac{d}{a} = 3 \cdot \frac{1}{2} \cdot \left(- 2\right) = - 3 , - \frac{b}{a} = 3 - 2 + \frac{1}{2} = \frac{3}{2} ,$

$\frac{c}{a} = 3 \left(- 2\right) + \left(- 2\right) \frac{1}{2} + \frac{1}{2} \left(3\right) = - 6 - 1 + \frac{3}{2} = - \frac{11}{2}$.

$\text{Or, } b = \frac{- 3 a}{2} , c = \frac{- 11 a}{2} , d = 3 a .$

$\text{Hence, } p \left(x\right) = a {x}^{3} - 3 \frac{a}{2} {x}^{2} - 11 \frac{a}{2} x + 3 a$

$= a \left({x}^{3} - \frac{3}{2} {x}^{2} - \frac{11}{2} x + 3\right) , w h e r e , a \in \mathbb{R} - \left\{0\right\}$

$\text{The Reqd. Poly. } p \left(x\right) = \frac{a}{2} \left(2 {x}^{3} - 3 {x}^{2} - 11 x + 6\right) , a \ne 0 , a \in \mathbb{R} .$