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

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

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Answer 1 · 2016-09-26T13:18:30

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

Explanation:

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

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

and $y = (x - a) (x - b) (x - c) is the polynomial$ $y=(x-a)(x-b)(x-c)" is the polynomial"$

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

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

$\Rightarrow y = (x - 3) (x + 2) (x - \frac{1}{2}) gives the polynomial$ $rArry=(x-3)(x+2)(x-1/2)" gives the polynomial"$

distribute the first 'pair' of brackets.

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

distributing and collecting like terms.

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

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

Answer 2 · 2016-09-26T13:25:37

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

Explanation:

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

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

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

= $2 (x - 3) (x + 2)) (x + \frac{1}{2})$ $2(x-3)(x+2))(x+1/2)$

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

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

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

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

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

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

Answer 3 · 2016-09-26T14:42:44

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

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 $(1)-(3)$ ,

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

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

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

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

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

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

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

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