Given:
#color(white)( (2x-3)/color(black)(2x-3))color(white)((d + e + f))/(")" color(white)(x)2x^2-7x+9)#
To find the first term of the quotient, divide the first term in the dividend by the first term in the divisor #(2x^2)/(2x) = x#. Write #x# in the quotient:
#color(white)( (2x-3)/color(black)(2x-3))(xcolor(white)( + e + f))/(")" color(white)(x)2x^2-7x+9)#
Multiply the divisor by the term in the quotient #x(2x-3)= 2x^2-3x#. Write the negative of this below the dividend and perform the addition:
#color(white)( (2x-3)/color(black)(2x-3))(xcolor(white)( + e + f))/(")" color(white)(x)2x^2-7x+9)#
#color(white)(".............")ul(-2x^2+3x)#
#color(white)(".......................")-4x+9#
To find the next term of the quotient, divide the first term of the results of the subtraction by the first term in the divisor #(-4x)/(2x) = -2#. Add #-2# to the quotient:
#color(white)( (2x-3)/color(black)(2x-3))(x-2color(white)( f))/(")" color(white)(x)2x^2-7x+9)#
#color(white)(".............")ul(-2x^2+3x)#
#color(white)(".......................")-4x+9#
Multiply the divisor by the new term in the quotient #-2(2x-3)= -4x^2+6#. Write the negative of this below the dividend and perform the addition:
#color(white)( (2x-3)/color(black)(2x-3))(x-2color(white)( f))/(")" color(white)(x)2x^2-7x+9)#
#color(white)(".............")ul(-2x^2+3x)#
#color(white)(".......................")-4x+9#
#color(white)("..........................")ul(4x-6)#
#color(white)("..................................")3#
Because power of the divisor is greater than the power of the results of the subtraction, we know that the results is the remainder. You can write the quotient plus the remainder as follows:
#x - 2+3/(2x-3)#
