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3
+ax
2
+3x−5 and q(x)=x
3
+x
2
−4x−a and the factor given is g(x)=x−1, therefore, by remainder theorem, the remainders are p(1) and q(1) respectively and thus,
p(1)=(2×1
3
)+(a×1
2
)+(3×1)−5=(2×1)+(a×1)+3−5=2+a−2=a
q(1)=1
3
+1
2
−(4×1)−a=1+1−4−a=−2−a
Now since it is given that both the polynomials p(x)=2x
3
+ax
2
+3x−5 and q(x)=x
3
+x
2
−4x−a leave the same remainder when divided by (x−1), therefore p(1)=q(1) that is:
a=−a−2
⇒a+a=−2
⇒2a=−2
⇒a=−
2
2
⇒a=−1
Hence, a=−1.
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