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) 2x³+x²−5x+2; ½, 1, −2 (ii) x³−4x²+5x−2; 2, 1, 1

(i) 2x³+x²−5x+2; ½, 1, −2; p(x)=2x³+x²−5x+2 (1)Zeroes for this polynomial are ½, 1, −2; Substitute the x=½ in equation (1) p(½)=2(½)³+(½)²−5(½)+2=1/4+1/4−5/2+2=0 Substitute the x=1 in equation (1) p(1)=2×1³+1²−5×1+2=2+1−5+2=0 Substitute the x=−2; in equation (1) p(−2;)=2(−2;)³+(−2;)²−5(−2;)+2=−16+4+10+2=0 Therefore, ½, 1, −2; are the zeroes of the given polynomial. Comparing the given polynomial with ax³+bx²+cx+d we obtain, a=2, b=1, c=−5, d=2 Let us assume α=½, β=1, γ=−2; Sum of the roots =α+β+γ=½+1−2=−½=−1/2=−b/a αβ+βγ+αγ=½+1(−2)+½(−2)=−5/2=c/a Product of the roots =αβγ=½×1×(−2)=−1=d/a Therefore, the relationship between the zeroes and coefficient are verified (ii) x³−4x²+5x−2; 2, 1, 1 p(x)=x³−4x²+5x−2; (1) Zeroes for this polynomial are 2, 1, 1 Substitute x=2 in equation (1) p(2)=2³−4×2²+5×2−2;=8−16+10−2;=0 Substitute x=1 in equation (1) p(1)=x³−4x²+5x−2;=1³−4(1)²+5(1)−2;=1−4+5−2;=0 Therefore, 2, 1, 1 are the zeroes of the given polynomial. Comparing the given polynomial with ax³+bx²+cx+d we obtain, a=1, b=−4, c=5, d=−2; Let us assume α=2, β=1, γ=1 Sum of the roots =α+β+γ=2+1+1=4=−−4/1=−b/a Multiplication of two zeroes taking two at a time=αβ+βγ+αγ=(2)(1)+(1)(1)+(2)(1)=5=5/1=c/a Product of the roots =αβγ=2×1×1=2=−−2/1=d/a Therefore, the relationship between the zeroes and coefficient are verified.