Question

please answer the following question​

Answer 1

Answer:

-c/d

Step-by-step explanation:

given polynomial f(X) = ax³+bx²+cx+d

zeroes of polynomial are A,B,Y

1/A+1/B+1/Y = (BY + AB+ AY)/ABY

from the formulas,

product of zeroes of a polynomial = -d/a

sum of product of each pair of roots(BY + AB+ AY)= c/a

so, 1/A+1/B+1/Y = (c/a)/(-d/a)

= -c/d

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Answer 2

Step-by-step explanation:

Given :-

A ,B , y are the zeroes of the Polynomial f(x) = ax³+bx²+cx+d.

To find :-

Find the value of (1/A)+(1/B)+(1/y)?

Solution :-

Given cubic polynomial f(x) = ax³+bx²+cx+d.

We know that

Sum of the zeroes = -b/a

A+ B + y = -b/a -------(1)

Sum of the product of the two zeroes taken at a time = c/a

=> AB + By+ yA = c/a --------(2)

Product of the zeroes = -d/a

=> ABy = -d/a ------------(3)

Now , We have to find the value of (1/A)+(1/B)+(1/y)

=>[ (1×B×y)+(1×A×y)+(1×A×B)]/ABy

=> (By+Ay+AB)/ ABy

From (2)&(3) then

=> (c/a)/(-d/a)

=> (c/a) × (-a/d)

=>(c×-a)/(a×d)

=> -ac / ad

=> - c/d

Therefore (1/A)+(1/B)+(1/y) = -c/d

Answer:-

The value of (1/A)+(1/B)+(1/y) for the given problem is -c/d

Used formulae:-

• The standard Cubic Polynomial is ax³+bx²+cx+d.

• Sum of the zeroes = -b/a

• Sum of the product of the two zeroes taken at a time = c/a

• Product of the zeroes = -d/a