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