How do you divide #(-4x^3-x^2+3x-4)/(x-3) #?

Question

I changed #(-4^3) rarr (-4x^3)#. Sorry if this is not what was intended, but it seemed more likely.

Avery Alexander · Accepted Answer

#-4x^2-13x-36-112/(x-3)#

#" "-4x^3-x^2+3x-4# #color(magenta)(-4x^2)(x-3) ->ul( -4x^3+12x^2) larr" Subtract"# #" "color(white)(..)0-13x^2+3x-4# #color(magenta)(-13x)(x-3)->" "ul(color(white)(..)-13x^2+39x)" "larr" Subtract"# #" "0color(white)(..)-36x-4# #color(magenta)(-36)(x-3)->" "color(white)(..)ul(-36x+108)" "larr" Subtract"# #" "0color(white)(..)color(magenta)(-112 larr" Remainder")#

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#color(magenta)(-4x^2-13x-36-112/(x-3))#

Jace Anderson · Answer

undefined To divide (-4x^3-x^2+3x-4) by (x-3), you can use long division. Here are the steps:

1. Divide the first term of the numerator (-4x^3) by the first term of the denominator (x). This gives -4x^2.
2. Multiply the entire denominator (x-3) by -4x^2, which gives -4x^3 + 12x^2.
3. Subtract this result from the numerator: (-4x^3 - x^2 + 3x - 4) - (-4x^3 + 12x^2) = -13x^2 + 3x - 4.
4. Bring down the next term from the numerator, which is 0x (since there is no x term).
5. Divide the first term of the new numerator (-13x^2) by the first term of the denominator (x). This gives -13x.
6. Multiply the entire denominator (x-3) by -13x, which gives -13x^2 + 39x.
7. Subtract this result from the new numerator: (-13x^2 + 3x - 4) - (-13x^2 + 39x) = -36x - 4.
8. Bring down the next term from the numerator, which is -36x.
9. Divide the first term of the new numerator (-36x) by the first term of the denominator (x). This gives -36.
10. Multiply the entire denominator (x-3) by -36, which gives -36x + 108.
11. Subtract this result from the new numerator: (-36x - 4) - (-36x + 108) = -112.
12. There are no more terms left in the numerator, so the division is complete.

The quotient is -4x^2 - 13x - 36, and the remainder is -112.