Question
The polynomials ax3+4x2+3x–4 and x3–4x+a leave the same remainder when divided by (x–3). Find the value of a.
Answer: Option A
:
A
Given polynomials:
p(x)=ax3+4x2+3x−4
q(x)=x3−4x+a
Using remainder theorm, the remainders when p(x)and q(x)are divided by (x−3)are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×3−4
p(3)=27a+41
q(3)=(3)3−4×3+a
q(3)=15+a
Given that the remainder is the same.
⇒p(3)=q(3)
27a+41=15+a
26a=26
a=1
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:
A
Given polynomials:
p(x)=ax3+4x2+3x−4
q(x)=x3−4x+a
Using remainder theorm, the remainders when p(x)and q(x)are divided by (x−3)are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×3−4
p(3)=27a+41
q(3)=(3)3−4×3+a
q(3)=15+a
Given that the remainder is the same.
⇒p(3)=q(3)
27a+41=15+a
26a=26
a=1
Was this answer helpful ?
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