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

3 Answers
Sep 26, 2016

Answer:

#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"#

Sep 26, 2016

Answer:

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#

Sep 26, 2016

Answer:

#"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.#