Question

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

Answer 1

`y=x^3-3/2x^2-11/2x+3`

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 2

The desired polynomial is `2x^3-x^2-13x-6`

Explanation:

A polynomial with zeros `p`, `q` and `r` can be written as `a(x-p)(x-q)(x-r)`

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

`2(x-3)(x-(-2))(x-(-1/2))`

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

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

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

= `(x-3)(2x^2+x+4x+2)`

= `x(2x^2+5x+2)-3(2x^2+5x+2)`

= `2x^3+5x^2+2x-6x^2-15x-6`

= `2x^3-x^2-13x-6`

Answer 3

`"The Reqd. Poly. "=2a(2x^3-3x^2-11x+6), a !=0, a in RR.`

Explanation:

We are given `3` real zeroes of a Poly., say `p(x)`.

Hence, `p(x)` has to be a cubic poly.

So, let, `p(x)=ax^3+bx^2+cx+d, ane0,`

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(x)=ax^3+bx^2+cx+d, ane0,` then,

`alpha+beta+gamma=-b/a..........(1)`,

`alphabeta+betagamma+gammaalpha=c/a...........(2)`,

`alphabetagamma=-d/a..........(3)`

In our case, we have, by `(1)-(3)`,

`-d/a=3*1/2*(-2)=-3, -b/a=3-2+1/2=3/2,`

`c/a=3(-2)+(-2)1/2+1/2(3)=-6-1+3/2=-11/2`.

`"Or, "b=(-3a)/2, c=(-11a)/2, d=3a.`

`"Hence, "p(x)=ax^3-3a/2x^2-11a/2x+3a`

`=a(x^3-3/2x^2-11/2x+3), where, a in RR-{0}`

`"The Reqd. Poly. "p(x)=a/2(2x^3-3x^2-11x+6), a !=0, a in RR.`