Let `alpha = 0, beta=0` and y be the zeros of the polynomial f(x)= ax3 + bx2 + cx + d Therefore `alpha + ß + y= (-text{coefficient of }X^2)/(text{coefficient of } x^3)` `= -(b/a)` `alpha+beta+y = -b/a` `0+0+y = -b/a` `y = - b/a` `\text{The value of} y - b/a`
Hence, the correct choice is `(c).` Page 2If two zeros x3 + x2 − 5x − 5 are \[\sqrt{5}\ \text{and} - \sqrt{5}\], then its third zero is Let `alpha = sqrt5` and `beta= -sqrt5` be the given zeros and y be the third zero of x3 + x2 − 5x − 5 = 0 then By using `alpha +beta + y = (-text{coefficient of }x^2)/(text{coefficient of } x^3)` `alpha + beta + y = (+(+1))/1` `alpha + beta + y = -1` By substituting `alpha = sqrt5` and `beta= -sqrt5` in `alpha +beta+y = -1` `cancel(sqrt5) - cancel(sqrt5) + y = -1` ` y = -1` Hence, the correct choice is`(b)` Concept: Relationship Between Zeroes and Coefficients of a Polynomial Is there an error in this question or solution? Page 3Given `alpha, beta,y` be the zeros of the polynomial x3 + 4x2 + x − 6 Product of the zeros = `(\text{Constant term })/(\text{Coefficient of}\x^3) = (-(-6))/1 =6` The value of Product of the zeros is 6.
Hence, the correct choice is `( c ).` > Suggest Corrections 0 Text Solution `(-b)/a` `b/a` `c/a` `(-d)/a` Answer : A Solution : Let ` alpha, 0, 0` be the zeros of `ax^(3) + bx^(2) + cx + d.` Then, <br> sum of zeros = `(-b)/a rArr alpha+0+0=(-b)/a rArr alpha = (-b)/a.` <br> Hence, the third zeros is `(-b)/a.` |