How do you divide #(-x^5+7x^3-x)div(x^3-x^2+1)# using long division?

Question

Madelyn Anderson · Accepted Answer

#=-x^2-x+6+(7x^2-6)/(x^3-x^2+1)#

For the polynomial divison we can see it as;

#(-x^5+7x^3-x) : (x^3-x^2+1) = #

So basically, what we want is to get rid of #(-x^5+7x^3-x)# here with something we can multiply on #(x^3-x^2+1)#.

We can start out with focusing on the first parts of the two, #(-x^5) : (x^3)#. So what do we need to multiply #(x^3)# with here in order to achieve #-x^5#? The answer is #-x^2#, because #x^3*(-x^2)=-x^5#.

So, #-x^2# will be our first part for the polynomial long divison. Now though, we cannot just stop at multiplying #-x^2# with the first part of #(x^3-x^2+1)#. We have to do it for each of the operands.

In that case, our first chosen operand will give us the result of; #x^3*(-x^2)-x^2*(-x^2)+1*(-x^2)#. Though there is one extra thing, there's always a #-# (minus) operator before the divison. So the notation would actually be something like,

#(-x^5+7x^3-x) : (x^3-x^2+1) = color(red)(-x^2)# #-(-x^5+x^4-x^2)#

Which will give us,

#(-x^4+7x^3+x^2-x):(x^3-x^2+1)#

A little notice here is that any operand that isn't taken out by the divison is carried on. That is untill we cannot do any divison. Meaning that we cannot find anything to multiply #(x^3-x^2+1)# with in order to take out any elements from the left side.

I will continue with the notation now,

#(-x^4+7x^3+x^2-x):(x^3-x^2+1) = color(red)(-x)# #-(-x^4+x^3-x)# #=>(6x^3+x^2) : (x^3-x^2+1)#

#(6x^3+x^2) : (x^3-x^2+1) = color(red)(6)# #-(6x^3-6x^2+6)# #=>(7x^2+6) : (x^3-x^2+1)#

It's a stop here. Because #(x^3-x^2+1)# contains a #x^3# and there is nothing on the left side that will need something #x^3#. We will then have our answer as;

#=-x^2-x+6+(7x^2-6)/(x^3-x^2+1)#

Elijah Allen · Answer

#-x^2-x+6+(7x^2-6)/(x^3-x^2+1)# Using place keepers of 0 value. Example: #0x^4# #color(white)("ddddddddddddddddd") -x^5+0x^4+7x^3+0x^2-x+0# #color(magenta)(-x^2)(x^3-x^2+1) ->color(white)("")ul(-x^5+color(white)(0)x^4+0x^3-x^2 larr" Subtract")# #color(white)("ddddddddddddddddddd")0color(white)("d")-x^4+7x^3+x^2-x+0 # #color(magenta)(-x)(x^3-x^2+1) ->color(white)("dddd.d")ul(-x^4+x^3+0x^2-xlarr" Subt")# #color(white)("dddddddddddddddddddddddd")0+6x^3 +x^2+0# #color(magenta)(6)(x^3-x^2+1) ->color(white)("ddddddddddd")ul(+6x^3-6x^2+6 larr" Subt")# #color(white)("dddddddddddddddddddddddddddd")color(magenta)(0+7x^2-6 larr" Remaind")# #color(magenta)( -x^2-x+6+(7x^2-6)/(x^3-x^2+1) )#

James Allen · Answer

undefined To divide (-x^5 + 7x^3 - x) by (x^3 - x^2 + 1) using long division:

1. Divide the leading term of the dividend by the leading term of the divisor: (-x^5) ÷ (x^3) = -x^2.
2. Multiply the entire divisor by the result from step 1: -x^2 * (x^3 - x^2 + 1) = -x^5 + x^4 - x^2.
3. Subtract the result from step 2 from the dividend: (-x^5 + 7x^3 - x) - (-x^5 + x^4 - x^2) = 7x^3 - x^4 + x^2 - x.
4. Repeat steps 1-3 with the new dividend: 7x^3 ÷ x^3 = 7.
5. Multiply the entire divisor by the result from step 4: 7 * (x^3 - x^2 + 1) = 7x^3 - 7x^2 + 7.
6. Subtract the result from step 5 from the dividend: (7x^3 - x^4 + x^2 - x) - (7x^3 - 7x^2 + 7) = -x^4 + 8x^2 - x - 7.
7. Repeat steps 1-3 with the new dividend: -x^4 ÷ x^3 = -x.
8. Multiply the entire divisor by the result from step 7: -x * (x^3 - x^2 + 1) = -x^4 + x^3 - x.
9. Subtract the result from step 8 from the dividend: (-x^4 + 8x^2 - x - 7) - (-x^4 + x^3 - x) = 8x^2 - x^3 - 7.
10. Repeat steps 1-3 with the new dividend: 8x^2 ÷ x^3 = 0 (as the degree of the dividend is less than the degree of the divisor).

The division process stops here. Therefore, the quotient is -x^2 + 7 - x, and the remainder is 8x^2 - x^3 - 7.