1) [tex]x^2=x\cdot x[/tex], and [tex]x(x+3)=x^2+3x[/tex]. Subtracting this from the numerator gives a remainder of
[tex](x^2-13x-48)-(x^2+3x)=-16x-48[/tex]
[tex]-16x=-16\cdot x[/tex], and [tex]-16(x+3)=-16x-48[/tex]. Subtracting this from the previous remainder gives a new remainder of
[tex](-16x-48)-(-16x-48)=0[/tex]
This means that
[tex]\dfrac{x^2-13x-48}{x+3}=x-16[/tex]
2) [tex]3x^3=3x^2\cdot x[/tex], and [tex]3x^2(x+2)=3x^3+6x^2[/tex]. Subtracting this from the numerator gives a remainder of
[tex](3x^3-x^2-7x+6)-(3x^3+6x^2)=-7x^2-7x+6[/tex]
[tex]-7x^2=-7x\cdot x[/tex], and [tex]-7x(x+2)=-7x^2-14x[/tex]. Subtracting this from the previous remainder gives a new remainder of
[tex](-7x^2-7x+6)-(-7x^2-14x)=7x+6[/tex]
[tex]7x=7\cdot x[/tex], and [tex]7(x+2)=7x+14[/tex]. Subtracting this from the previous remainder gives a new remainder of
[tex](7x+6)-(7x+14)=-8[/tex]
This means that
[tex]\dfrac{3x^3-x^2-7x+6}{x+2}=3x^2-7x+7-\dfrac8{x+2}[/tex]
3) [tex]x+2[/tex] will be a factor of [tex]x^3+3x^2-10x-24[/tex] if dividing the latter by [tex]x+2[/tex] leaves a remainder of 0.
[tex]x^3=x^2\cdot x[/tex], and [tex]x^2(x+2)=x^3+2x^2[/tex]. Subtracting this from the numerator gives a remainder of
[tex](x^3+3x^2-10x-24)-(x^3+2x^2)=x^2-10x-24[/tex]
[tex]x^2=x\cdot x[/tex], and [tex]x(x+2)=x^2+2x[/tex]. Subtracting this from the previous remainder gives a new remainder of
[tex](x^2-10x-24)-(x^2+2x)=-12x-24[/tex]
[tex]-12x=-12\cdot x[/tex], and [tex]-12(x+2)=-12x-24[/tex]. Subtracting this from the previous remainder gives a new remainder of
[tex](-12x-24)-(-12x-24)=0[/tex]
and since the remainder is 0, [tex]x+2[/tex] is indeed a factor of [tex]x^3+3x^2-10x-24[/tex].
