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 + aIt 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=4Now 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) + a64a + 48-3 = 128 - 20 + a64a - a = 108 - 4563 a = 63a = 1