Question

因式分解问题，求回答！！

若x^2-8x+13=0，求（x^4-6x^3-2x^2+18x+23）\（x^2-8x+15）

Answer 1

x^2-8x+13=0
x^2-8x=-13

(x^4-6x^3-2x^2+18x+23)/(x^2-8x+15)
=(x^4-6x^3-2x^2+18x+23)/(-13+15)
=(x^4-6x^3-2x^2+18x+23)/2
=(x^4-8x^3+2x^3-2x^2+18x+23)/2
=[x^2(x^2-8x)+2x^3-2x^2+18x+23]/2
=[-13x^2+2x^3-2x^2+18x+23]/2
=[2x^3-15x^2+18x+23]/2
=[2x^3-16x^2+x^2+18x+23]/2
=[2x(x^2-8x)+x^2+18x+23]/2
=[-26x+x^2+18x+23]/2
=[x^2-8x+23]/2
=(-13+23)/2
=10/2
=5

追问

请问
=[-26x+x^2+18x+23]/2
这一步中-26从何而来？
谢谢

追答

x^2-8x=-13
[2x(x^2-8x)+x^2+18x+23]/2
=[2x*(-13)+x^2+18x+23]/2
=[-26x+x^2+18x+23]/2