The optional chapter of polynomials , how to solve the question of exercise 2.4 of question 1. I don’t know the solution of this question. How can i solve easily by simple way, the question is Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (i) 2×3+x2-5x+2; -1/2, 1, -2 (ii) x3-4×2+5x-2 ;2, 1, 1

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# Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: (i) 2×3+x2-5x+2; -1/2, 1, -2 (ii) x3-4×2+5x-2 ;2, 1, 1 Q.1

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(i) 2x

^{3}+x^{2}-5x+2; -1/2, 1, -2Solution:

Given, p(x) = 2x

^{3}+x^{2}-5x+2And zeroes for p(x) are = 1/2, 1, -2

âˆ´ p(1/2) = 2(1/2)

^{3}+(1/2)^{2}-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0p(1) = 2(1)

^{3}+(1)^{2}-5(1)+2 = 0p(-2) = 2(-2)

^{3}+(-2)^{2}-5(-2)+2 = 0Hence, proved 1/2, 1, -2 are the zeroes of 2x

^{3}+x^{2}-5x+2.Now, comparing the given polynomial with general expression, we get;

âˆ´ ax

^{3}+bx^{2}+cx+d = 2x^{3}+x^{2}-5x+2a=2, b=1, c= -5 and d = 2

As we know, if Î±, Î², Î³ are the zeroes of the cubic polynomial ax

^{3}+bx^{2}+cx+d , then;Î± +Î²+Î³ = â€“b/a

Î±Î²+Î²Î³+Î³Î± = c/a

Î± Î²Î³ = â€“ d/a.

Therefore, putting the values of zeroes of the polynomial,

Î±+Î²+Î³ = Â½+1+(-2) = -1/2 = â€“b/a

Î±Î²+Î²Î³+Î³Î± = (1/2Ã—1)+(1 Ã—-2)+(-2Ã—1/2) = -5/2 = c/a

Î± Î² Î³ = Â½Ã—1Ã—(-2) = -2/2 = -d/a

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

(ii) x

^{3}-4x^{2}+5x-2 ;2, 1, 1Solution:

Given, p(x) = x

^{3}-4x^{2}+5x-2And zeroes for p(x) are 2,1,1.

âˆ´ p(2)= 2

^{3}-4(2)^{2}+5(2)-2 = 0p(1) = 1

^{3}-(4Ã—1^{2 })+(5Ã—1)-2 = 0Hence proved, 2, 1, 1 are the zeroes of x

^{3}-4x^{2}+5x-2Now, comparing the given polynomial with general expression, we get;

âˆ´ ax

^{3}+bx^{2}+cx+d = x^{3}-4x^{2}+5x-2a = 1, b = -4, c = 5 and d = -2

As we know, if Î±, Î², Î³ are the zeroes of the cubic polynomial ax

^{3}+bx^{2}+cx+d , then;Î± + Î² + Î³ = â€“b/a

Î±Î² + Î²Î³ + Î³Î± = c/a

Î± Î² Î³ = â€“ d/a.

Therefore, putting the values of zeroes of the polynomial,

Î± +Î²+Î³ = 2+1+1 = 4 = -(-4)/1 = â€“b/a

Î±Î²+Î²Î³+Î³Î± = 2Ã—1+1Ã—1+1Ã—2 = 5 = 5/1= c/a

Î±Î²Î³ = 2Ã—1Ã—1 = 2 = -(-2)/1 = -d/a

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