# If polynomials ax^{3} + 3x^{2} - 3 and 2x^{3} - 5x + a when divided by (x - 4) leave the remainders as R1 and R2 respectively. Find the values of a in 2R1 - R2=0

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 ) = ax^{3} + 3x^{2} - 3 and g( x ) = 2x^{3} - 5x + a

Given that f( x ) and g( x ) when divided by x - 4 leaves the remainders R_{1} and R_{2} respectively.

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

For f( 4 ) = ax^{3} + 3x^{2} - 3

= a × ( 4 )^{3} + 3 × ( 4 )^{2} - 3 = 64 a + 48 - 3

= 64 a + 45 = R_{1} (eq 1)

For g( 4 ) = 2x^{3} - 5x + a

= 2 × ( 4 )^{3} - 5 × ( 4 ) + a = 128 - 20 + a

= 108 + a = R_{2} (eq 2)

Given that 2R_{1} - R_{2} = 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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