InterviewSolution
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InterviewSolution
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All possible roots of polynomial with integral coefficients can be identified by "RATIONAL ROOT TEST"according to rational root test if a plynomial ltbr. a_(n)p^(n)+a_(n-1)p^(n-1)q+a_(n-2)p^(n-2)q^(2)+…..+a_(2)p^(2)q^(n-2)+a_(1)pq^(n-1)+a_(0)q^(n)=0 Every term in above equation except possible the last one is divisible by p hence p should also divide a_(0)q^(n), since q and p are relatively prime p must divide a_(0). Similarly q also divides a_(n). Now consider an equation 6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0 and answer the following questions Q. if given equation and equation x^(3)+ax^(2)+ax+1=0 have two roots common then possible value(s) of is/are? |
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Answer» `0` `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` rational roots of equation `x=+-1,+-(1)/(2),+-(1)/(2),+-(1)/(6)` are possible `x=(1)/(6)` satisfies the equation `6x^(5)-x^(4)-18X^(4)+3x^(3)-12x^(3)+2x^(2)-18x^(2)+3x+6x-1=0` `(6x-1)(x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1=0` `x^(2)+(1)/(x^(2))-3(x+(1)/(x))-2=0` `(x+(1)/(x))^(2)-3(x+(1)/(x))-4=0` `t^(2)-3t-4=0` `(t-4)(t+1)=0` `x+(1)/(x)=4` or `x+(1)/(x)+1=0` `x^(2)-4x+1=0` `x^(2)+x+1=0` sum of REAL roots will be `4+(1)/(6)=(25)/(6)` for rational values of a roots of CUBIC `x^(2)+ax^(2)+ax+1=0` will be in conjugate pair so either `x^(2)-4x+1=0` or `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+1=0` if `x^(2)-4x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=-3` and `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=2` |
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