A) 2x
B) -2X
C) x
D) -x
Correct Answer: D
Solution :
[d] We are given that \[p(-a)=a\text{ }and\text{ }p(a)=-a\] |
[When a polynomial f(x) is divided by x-a, remainder is f(a)], |
Let the remainder, when p(x) is divided by\[{{x}^{2}}-{{a}^{2}}\], be Ax+B. Then,. |
\[p(x)=Q(x)({{x}^{2}}-{{a}^{2}})+Ax+B\] (1) |
Where Q(x) is the quotient. Putting x=a and -a in (1), we get |
\[p(a)=0+Aa+B\Rightarrow -a=Aa+B\] (2) |
And \[p(-a)=0-aA+B\Rightarrow a=-aA+B\] (3) |
Solving (2) and (3), we get |
B=0 and A=-1 |
Hence, the required remainder is-x. |
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