Question

26) If the polynomials 2x3 + ax2 + 3x - 5 and x3 + x2 - 4x + a leave the same remainder when divided by
X-2. Find the value of a.
​

Answer 1

Answers Let p(x)=2x

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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