Question

what Will be the remainder when 6x^5+4x^4-27x^3-7x^2+27x+3/2 is divided by (2x^2-3)?​

Answer 1

Answer:

0

Step-by-step explanation:

let p(x) = 6x^5+4x^4-27x^3-7x^2+27x+3/2

when divided by (2x^2-3)

the remainder will be p (±√3/√2)

p (±√3/√2) = 6 (±√3/√2)∧5 + 4 (±√3/√2)∧4 - 27 (±√3/√2)³ - 7 (±√3/√2)²

+ 27 (±√3/√2) + 3/2

p (±√3/√2) = 6 ×9/4(±√3/√2) + 4 × 9/4 - 27×3/2 (±√3/√2) - 7×3/2

+ 27 (±√3/√2) + 3/2

p (±√3/√2)  = (±√3/√2) ( 27/2 - 81/2 + 27 ) + 9 - 21/2 + 3/2

p (±√3/√2)  = (±√3/√2) ( 27/2 - 81/2 + 27 ) + 9 - 21/2 + 3/2

p (±√3/√2) = 0

Remainder = 0