**Question 1.** **Asked on :**12 January 2019:10:45:37 AM

If the polynomials ax^{3}+3x^{2}-13 and 2x^{3}-5x+a , are divided by x+2 if the remainder in each case is the same, then find the value of 'a'

*-Added by Akki chauhan*

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

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*-Answered by Himanshi Verma* On 20 January 2019:05:28:21 PM

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

*-Answered by Himanshi Verma* On 20 January 2019:05:29:25 PM

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

*-Answered by Master Purushottam* On 22 August 2019:09:12:19 PM

**Answer:**

When p(x) is divided by (x-2),

By remainder theorem,

Remainder= p(2)

p(2)=a*2^3 + 3*2^2 - 13

=a*8+3*4-13

=8a+12-13 = 8a-1

p(2) = 2*2^3-5*2+a

=2*8-10+a

=16-10+a = 6+a

Since the remainders of ax^3+3x^2-13 and 2x^3-5x+a are same when divided by x-2,

8a-1 = 6+a

8a-6+a=1

8a+8=1-6=-5

8a=-5-8 =-13

a=-13/8

*-Answered by Shivang Gupta* On 25 August 2019:05:08:54 PM

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