The polynomial ax^3+3x^2-3 and 2x3-5x+a when divided by x-4 leaves the same remainder

A polynomial is an algebraic expression that has constants, variables, and coefficients with a point where the value of the polynomial becomes zero as a whole.

Answer: The value of a in 2 R1 - R2 = 0 is 18 / 127.

Here's the step-by-step solution.

Explanation:

Let f[ x ] = ax3 + 3x2 - 3 and g[ x ] = 2x3 - 5x + a

Given that f[ x ] and g[ x ] when divided by x - 4 leaves the remainders R1 and R2 respectively.

By remainder theorem, substituting the value x = 4 in both f[ x ] and g[ x ], we get remainders.

For f[ 4 ] = ax3 + 3x2 - 3

= a × [ 4 ]3 + 3 × [ 4 ]2 - 3 = 64 a + 48 - 3

= 64 a + 45 = R1    [eq 1]

For g[ 4 ] = 2x3 - 5x + a

=  2 × [ 4 ]3 - 5 × [ 4 ] + a = 128 - 20 + a

= 108 + a = R2  [eq 2]

Given that 2R1 - R2 = 0

Therefore, from eq 1 and eq 2

2 [ 64 a + 45 ] - [ 108 + a ] = 0

⇒ 128 a + 90 - 108 - a = 0

⇒ 127 a - 18 = 0

⇒ 127 a = 18 

⇒ a = 18 / 127

Thus, the value of a in 2 R1 - R2 = 0 is  18 / 127.

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Solution

Given polynomials

P[

x1 ]= ax3 +3x2 - 3 and
p[
x2 ]= 2x3 - 5x + a

It is also given that these two polynomials leave the same remainder when divided by [x - 4].

i.e., [x-4] is the zero of the polynomial so, x=4

Now put the value of 'x' in the polynomials,

As both the Eq. have the same remainder so,

p[x1 ]=p[x2 ]

a[43 ] + 3[42 ] - 3 = 2[43 ] - 5[4] + a

64a + 48-3 = 128 - 20 + a

64a - a = 108 - 45

63 a = 63

a = 1


Solve

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