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Question -

Verify that the numbers given alongside of thecubic polynomials below are their zeroes. Also verify the relationship betweenthe zeroes and the coefficients in each case:

(i) 2x3+x2-5x+2; -1/2,1, -2               (ii) x3-4x2+5x-2 ;2, 1, 1



Answer -

(i) 2x3+x2-5x+2; -1/2,1, -2

Solution:

Given, p(x) 2x3+x2-5x+2

And zeroes for p(x) are = 1/2, 1, -2

  p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2

             = (1/4)+(1/4)-(5/2)+2

              = 0

     p(1)= 2(1)3+(1)2-5(1)+2  

        = 0

      p(-2)= 2(-2)3+(-2)2-5(-2)+2

         = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.

Now, comparing the given polynomial withgeneral expression, we get;

ax3+bx2+cx+d

 = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of thepolynomial,

α+β+γ = ½+1+(-2)  

            = -1/2

            = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2)

                  = -5/2

                  = c/a

α β γ = ½×1×(-2)

         = -2/2

         = -d/a

Hence, the relationship between the zeroes andthe coefficients are satisfied.

(ii) x3-4x2+5x-2 ;2, 1, 1

Solution:

Given, p(x) = x3-4x2+5x-2

And zeroes for p(x) are 2,1,1.

p(2)= 23-4(2)2+5(2)-2

         = 0

p(1) = 13-(4×1)+(5×1)-2

        = 0

Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2

Now, comparing the given polynomial withgeneral expression, we get;

ax3+bx2+cx+d = x3-4x2+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of thecubic polynomial ax3+bx2+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of thepolynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes andthe coefficients are satisfied.

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