We have been asked to determine the value of ‘a’ if the division of x3+5x2−ax+6 by (x−1) leaves the remainder 2a, using remainder theorem
ML Aggarwal(avichal publication) Factorisation chapter 6, question no 6
Rajan@2021Guru
Using remainder theorem, find the value of ‘a’ if the division of x 3 +5x 2 −ax+6 by (x−1) leaves the remainder 2a.
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Let us assume
x−1=0⇒x=1
By remainder theorem, when f(x) is divide by (x−1),
Remainder r=f(1)
⇒2a=(1)3+5(1)2−a(1)+6[∵r=2a(Given)]
⇒2a=1+5−a+6
⇒2a+a=12
⇒3a=12
⇒a=12/3
⇒a=4
Therefore, the value of a is 4.