Question

if The polynomial (2x³+ax²+3x-5)
and (x³+x²-2x+a) leave the same reminder when divided by (x-2) find the value of a. Also find the reminder in each case

Answer 1

hay!!

Dear user -

Let f(x) =2x³+ax²+3x -5 and (x³+x²-2x+a).
$let \: f(x) \: is \: divided \: by \\ \: (x - 2) \: reminder \: = f(2)$

$when \: g(x) \: is \: \\ divided \: by \: (x - 2) \\ reminder \: = g(2)$

Now, f(2)=> (2×2³+a×2²+3×2-5)=(17+4a).

and g(2)=> (2³+2²-2×2+a)=(8a+a)

17+4a=8+a

=> 3a=-9

=> a=-3

Hence, a=>-3

Reminder in each =(8-3)=5

I hope it's help you

Answer 2

Answer:

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.

Step-by-step explanation: