Answer:
[tex](x^3-11x^2+ 22x + 40)\div (x-5)=(x^2-6x-8)[/tex]
Step-by-step explanation:
When dividing [tex](x^3-11x^2+ 22x + 40)[/tex] by [tex](x-5),[/tex]
1. Multiply [tex](x-5)[/tex] by [tex]x^2[/tex] and subtract the result from [tex](x^3-11x^2+ 22x + 40)[/tex] to eliminate [tex]x^3:[/tex]
[tex](x^3-11x^2+ 22x + 40)-x^2(x-5)\\ \\=x^3-11x^2+22x+40-x^2+5x^2\\ \\=-6x^2+22x+40[/tex]
2. Multiply [tex](x-5)[/tex] by [tex]-6x[/tex] and subtract the result from [tex]-6x^2+22x+40[/tex] to eliminate [tex]-6x^2:[/tex]
[tex]-6x^2+22x+40-(-6x)(x-5)\\ \\=-6x^2+22x+40+6x^2-30x\\ \\=-8x+40[/tex]
3. Multiply [tex](x-5)[/tex] by [tex]-8[/tex] and subtract the result from [tex]-8x+40[/tex] to eliminate [tex]-8x:[/tex]
[tex]-8x+40-(-8)(x-5)\\ \\=-8x+40+8x-40\\ \\=0[/tex]
So,
[tex](x^3-11x^2+ 22x + 40)\div (x-5)=(x^2-6x-8)[/tex]
