How do you divide #(2x^2-3x+2)/(x - 1)#?

Question

Eva Adamson · Accepted Answer

#2x-1# and remainder of #(-3)/(x-1)# # color(white)(.............)ul(2x-1)# #color(white)(aa)x-1##|##2x^2-3x-2# #color(white)(..............)ul(2x^2-2x)# #color(white)(........................)-x-2# #color(white)(.........................)ul(-x+1)# #color(white)(..............................)-3# #color(magenta)((2x^2-3x-2) / (x-1) = 2x-1# and remainder#color(magenta)((-3)/(x-1)#

Nathan Axel · Answer

#2x^2-3x-2 = (x-1)(2x-1) - 3#

or

#(2x^2-3x-2)/(x-1) = 2x-1 - 3/(x-1)# In each section of the 'long division', the values on the quotient depend on what factor of the highest degree of the divisor can fit into the highest degree of the dividend. For example, #x# fits #2x# times into #2x^2#, #x# fits -1 times into #-x# #{: (,,,2x,-1), (,,"---","---","---"), (x-1,")",2x^2, -3x, -2), (,,2x^2,-2x,), (,,"---","---",), (,,,-x,-2), (,,,-x,+1), (,,"---","---","---"), (,,,,-3) :}# Therfore, #(2x^2-3x-2)/(x-1) = 2x-1 - 3/(x-1)# #2x^2-3x-2 = (x-1)(2x-1) - 3#

Savannah Anderson · Answer

undefined To divide (2x^2-3x+2) by (x - 1), we can use polynomial long division.

First, divide the highest degree term of the numerator (2x^2) by the highest degree term of the denominator (x). This gives us 2x as the quotient.

Next, multiply the entire denominator (x - 1) by the quotient (2x), which gives us 2x^2 - 2x.

Subtract this result from the numerator (2x^2 - 3x + 2) to get -x + 2.

Now, bring down the next term from the numerator (-x) and divide it by the highest degree term of the denominator (x). This gives us -1 as the next term of the quotient.

Multiply the entire denominator (x - 1) by the new quotient (-1), which gives us -x + 1.

Subtract this result from the previous remainder (-x + 2) to get 1.

Since the degree of the remainder (1) is less than the degree of the denominator (x - 1), we have finished the division.

Therefore, the quotient is 2x - 1 and the remainder is 1.