The graph of polynomial f(x) =  ax^{3} +b x^{2} +cx +d cut x-axis at (-1,0), (1,0) and y axis at (0,2). then the sum of its roots is?
This question was asked in exam of VMC. My answer was 0(zero). Am i ri8 plzz explain....!

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0 us definitely wrong because there exists another real because it is a cubic
*is
x^3-x is a cubic equation which has roots - 1, -1, 0. And the sum of its roots is 0. So it is possible to have zero as a root of equation.
does that cubic satisfy the point (0, 2) ?

Answers

2015-04-13T23:30:38+05:30

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P(x) = a x³ +  b x² + c x + d              let  a ≠ 0 

  P(x) cuts x axis at  (-1, 0)  =>  P(-1) = 0   
                            (1, 0)  =>  P(1) = 0        
                             (0, 2)  => P(0) = 2      
 
P(-1) = -a + b - c + d = 0     --- (1)
P(1) = a + b + c + d = 0    -- (2)
P(0) =  d  = 2      =>

add (1) and (2)  and substitute value of d:    b + d = 0  => b = -2   
   Hence,  a + c = 0  =>  a = - c.

P(x)  = a  (x³ +  b/a  x² + c/a  x + d/a)

Sum of the roots is =  -b/a  = - 1 * coefficient of x²   so that is   2/a  or  -2/c

also, the product of the three roots is  = d/a   : ie.,  2/a

since the sum is = 2/a
we have two roots given ,:  -1 and + 1  =>  third root is :  d/a  = 2/a



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