above is the link for the sollution and u need to explain me few steps where i was it was decided that other two zeroes are the zeroes of ax^2+bx +c

f(x)=ax^3+bx^2+cx = x(ax^2+bx+c)

it is said that the other two zeroes are the zeroes of quadratic in standard form.explain me this point


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Yes, the early part of the solution shows that constant term d must be zero so that f(0)=0.

And that leads to the factorisation x(ax^2+bx+c) and this expression=0 from which x=0 (which we are given anyway) or the quadratic ax^2+bx+c=0. There are various ways of solving this. The formula gives us two roots which could be complex. However, we can assume that the factorisation (x-A)(x-B) applies, where A and B are the roots, the zeroes. But, we have to divide through by a first so that we can have this factorised expression: if ax^2+bx+c=0 then x^2+bx/a+c/a=0=x^2-(A+B)x+AB.

We can now equate coefficients: (A+B)=-b/a and AB=c/a. Therefore the product of the zeroes, or roots, is c/a.

by Top Rated User (766k points)
good morning sir

one more explanation can u tell me how u got -(a+b)x.please show me this step
good morning sir

one more explanation can u tell me how u got -(a+b)x.please show me this step

When you expand (x-A)(x-B) you get x^2-Bx-Ax+AB. The two x terms combine as x(-B-A) which is the same as -x(A+B). The question only asks for the product AB in terms of the coefficients a, b and c, so we don't need to work out what the sum of the zeroes is.

thanq very much sir . u helped me

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