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

##### 3 Answers

#### Answer:

#### Explanation:

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

Then

#(x-a),(x-b)" and " (x-c)" are the factors"# and

#y=(x-a)(x-b)(x-c)" is the polynomial"# Here the zeros are

#x=3,x=-2" and " x=1/2#

#rArr(x-3),(x+2)" and " (x-1/2)" are the factors"#

#rArry=(x-3)(x+2)(x-1/2)" gives the polynomial"# distribute the first 'pair' of brackets.

#=(x^2-x-6)(x-1/2)# distributing and collecting like terms.

#=x^3-1/2x^2-x^2+1/2x-6x+3#

#rArry=x^3-3/2x^2-11/2x+3" is the polynomial"#

#### Answer:

The desired polynomial is

#### Explanation:

A polynomial with zeros

Hence a polynomial with zeros

=

=

=

=

=

=

=

#### Answer:

#### Explanation:

We are given

Hence, **cubic** poly.

So, let,

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,

zeroes of a Cubic Poly.

In our case, we have, by