Question

verify the division algorithm 6 x cube + 2 x square - 4 x + 3 is equal to 3 x square - 2 X + 1 into 2 X + 2 + - 2 X + 1​

Answer 1

Given that :

$dividend \: = \: {6x}^{3} + {2x}^{2} - 4x + 3 \\ divisor \: \: \: \: \: = {3x}^{2} - 2x + 1 \\ quotient \: \: = 2x + 2 \\ remainder = - 2x + 1$

To prove :

Dividend = Divisor × Quotient + Remainder

${6x}^{3} + {2x}^{2} - 4x + 3 = ( {3x}^{2} - 2x + 1)(2x + 2) + ( - 2)x + 1 \\ {6x}^{3} + {2x }^{2} - 4x + 3 =(2x(3 {3x}^{2} - 2x + 1 )+ 2( {3x}^{2} - 2x + 1) \\ {6x}^{3} + {2x}^{2} - 4x + 3 = {6x}^{3} - {4x}^{2} + 2x + {6x}^{2} - 4x + 2 \\ {6x}^{3} + {2x}^{2} - 4x + 3 = {6x}^{3} + {2x}^{2} - 2x + 2 \: \: \\ \\ \: therefore \: \: . \\ \\ {6x}^{3} + {2x}^{2} - 4x + 3 \: \: is \: not \: equals \: to \: {6x}^{3} + {2x}^{2} - 2x + 2$

SO,,,,

Dividend ≠ Divisor × Quotient + Remainder