According to remainder theorem if f(x)f(x) is divided by (x-a)(x−a), then remainder is f(a)f(a). Therefore if (x-a)(x−a) is a factor of f(x)f(x), f(a)=0f(a)=0.
Now coming to questions raised by you solution is given seriatim.
(a) As f(x)=x^n-a^nf(x)=xn−an, dividing by x-ax−a gives a remainder f(a)=a^n-a^n=0f(a)=an−an=0. Hence x^n-a^nxn−an is divisible by x-ax−a.
(b.I) As f(x)=x^n+a^nf(x)=xn+an, dividing by x+ax+a gives a remainder f(a)=(-a)^n+a^n=(-1)^na^n+a^nf(a)=(−a)n+an=(−1)nan+an. This will be 00 only if nn is odd. Hence the condition for x+ax+a to be a factor of x^n-a^nxn−an is nn is odd.
(b.II) As f(x)=x^n-a^nf(x)=xn−an, dividing by x+ax+a gives a remainder f(a)=(-a)^n-a^n=(-1)^na^n-a^nf(a)=(−a)n−an=(−1)nan−an. This will be 00 only if nn is even. Hence condition for x+ax+a to be a factor of x^n-a^nxn−an is nn is even.
3 c As f(x)=2x^3+x^2-5x+2f(x)=2x3+x2−5x+2, the factors are of the type x-ax−a, where aa is a factor of 2/222 (p/qpq where pp is constant term and qq is coefficient of highest power of xx) i.e. +-1±1 or +-2±2 or +-1/2±12. As f(1)=0f(1)=0, f(2)=0f(2)=0 and f(1/2)=0f(12)=0, hence factors are x-1x−1, x-2x−2 and 2x-12x−1 i.e. 2x^3+x^2-5x+2=(x-1)(x-2)(2x-1)2x3+x2−5x+2=(x−1)(x−2)(2x−1).
3.7 When P(x)P(x) is divided by (x-1)(x−1), remainder is 22, hence P(1)=2P(1)=2 and as when P(x)P(x) is divided by (x-2)(x−2), remainder is 33, hence P(2)=3P(2)=3.
(a) As P(x)=(x-1)(x-2)Q(x)+ax+bP(x)=(x−1)(x−2)Q(x)+ax+b,
P(1)=2P(1)=2 gives us a+b=2a+b=2 and P(2)=3P(2)=3 gives us 2a+b=32a+b=3.
Subtracting former from latter, we get a=1a=1 and hence b=1b=1
(b.I) As P(x)=(x-1)(x-2)Q(x)+ax+bP(x)=(x−1)(x−2)Q(x)+ax+b and P(x)P(x) is a cubic polynomial, with coefficient of x^3x3 as 11, it is of the type
P(x)=(x-1)(x-2)(x-k)+x+1P(x)=(x−1)(x−2)(x−k)+x+1. Now as -1−1 is a solution to P(x)=0P(x)=0, we have
P(-1)=(-1-1)(-1-2)(-1-k)-1+1=-6-6k=0P(−1)=(−1−1)(−1−2)(−1−k)−1+1=−6−6k=0
i.e. k=-1k=−1 and P(x)=(x-1)(x-2)(x+1)+x+1P(x)=(x−1)(x−2)(x+1)+x+1 or
P(x)=x^3-2x^2-4x-1P(x)=x3−2x2−4x−1
(b.II) It is apparent that as P(x)=(x-1)(x-2)(x+1)+x+1P(x)=(x−1)(x−2)(x+1)+x+1, (x+1)(x+1) is a factor of P(x)P(x) and
P(x)=(x-1)(x-2)(x+1)+x+1P(x)=(x−1)(x−2)(x+1)+x+1
= (x+1)(x^2-3x+2+1)=(x+1)(x^2-3x+3)(x+1)(x2−3x+2+1)=(x+1)(x2−3x+3)
As discriminant in x^2-3x+3 is 3^2-4xx1xx3=9-12=-3,
there is no other real factor.
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