How to long divide #(x^3-2x^2-4x-4)/(x^2+x-2)#?

Question

Benjamin Achtenberg · Accepted Answer

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+4/(x+2)+3/(x-1)#

#(x^3-2x^2-4x-4)/(x^2+x-2)#

By long division,

Hence,

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+color(green)((x-10)/(x^2+x-2)#

Then, let #a# and #b# be unknowns,

#color(green)((x-10)/(x^2+x-2))=(x-10)/((x+2)(x-1))#
#color(white)(xxxxxx//x)=a/(x+2)+b/(x-1)#

Multiply throughout by #x^2+x-2#,

#x-10=a(x-1)+b(x+2)#

When #color(red)(x=1#,

#color(red)(1)-10=a(color(red)(1)-1)+b(color(red)(1)+2)#
#color(white)(xxx)3b=-9#
#color(white)(xxx3)b=-3#

When #color(blue)(x=-2#,

#color(blue)(-2)-10=a(color(blue)(-2)-1)+b(color(blue)(-2)+2)#
#color(white)(....)-3a=-12#
#color(white)(....-3)a=4#

Hence, substitute #a=4# and #b=-3#,

#(x^3-2x^2-4x-4)/(x^2+x-2)=x-3+4/(x+2)+3/(x-1)#

Lily Anderson · Answer

undefined To long divide $(x^3-2x^2-4x-4)$ by $(x^2+x-2)$, you follow these steps:

1. Divide the leading term of the numerator by the leading term of the denominator to get the first term of the quotient.
2. Multiply the entire denominator by the first term of the quotient obtained in step 1.
3. Subtract the result from the numerator.
4. Repeat steps 1-3 with the result obtained in step 3 until you can't go any further.

Here are the steps in detail:

1. Divide $x^3$ by $x^2$ to get $x$, which is the first term of the quotient.
2. Multiply $(x^2+x-2)$ by $x$ to get $x^3+x^2-2x$.
3. Subtract $(x^3+x^2-2x)$ from $x^3-2x^2-4x-4$ to get $-3x^2-2x-4$.
4. Now, repeat the process with $-3x^2-2x-4$ as the new numerator.

Continuing in this manner, you can proceed with the long division process until you have reached the desired degree or until the remaining polynomial cannot be further divided by the divisor.