How does λ − 4 divided by λ^3 − 8λ^2 + 17λ − 4 equal λ^2 − 4λ + 1?

[nicholaskong100]

How does λ − 4 divided by λ^3 − 8λ^2 + 17λ − 4 equal λ^2 − 4λ + 1?​

 

[skeeter]

How does λ − 4 divided by λ^3 − 8λ^2 + 17λ − 4 equal λ^2 − 4λ + 1?​

I think you mean the other way around, the cubic polynomial divided by the linear one … ?

4] 1  -8  17  -4
       4 -16   4
------------------
   1  -4   1   0

note the depressed polynomial coefficients show the quotient as [imath]\lambda^2 -4\lambda +1[/imath]

 

[Steven G]

As pointed out you have the division backward.
You really should know how to check a division problem. If a/b=c, then a*c=b.
Just verify that (λ²−4λ+1)(λ-4)=λ³-8λ²+17λ-4

 

[nasi112]

By doing the long division, you will see how.

The question should be, why \(\displaystyle (\lambda - 4)\) is a factor of \(\displaystyle \lambda^3 - 8\lambda^2 + 17\lambda - 4\)?

Plug a number in \(\displaystyle \lambda\), say \(\displaystyle n\), if the result \(\displaystyle = 0\), then \(\displaystyle (\lambda - n)\) is a factor.

Testing...

\(\displaystyle 4^3 - 8(4)^2 + 17(4) - 4 = 0\)

 

[NotJeffM]

If you are asking why that all works and are not satisfied with memorizing the mechanics of division

[math] \lambda^3 - 8 \lambda^2 + 17 \lambda - 4 = \\ \lambda^3 - 4 \lambda^2 - 4 \lambda^2 + 17 \lambda - 4 =\\ \lambda^2( \lambda - 4) - 4 \lambda^2 + 16 \lambda + \lambda - 4 = \\ \lambda^2( \lambda - 4) - 4\lambda (\lambda - 4) + 1(\lambda - 4) = (\lambda - 4)(\lambda^2 - 4 \lambda + 1).\\ \therefore \ \dfrac{\lambda^3 - 8 \lambda^2 + 17 \lambda - 4}{\lambda - 4} = \dfrac{\cancel{(\lambda - 4)}( \lambda^2 - 4 \lambda + 1)}{\cancel {(\lambda - 4)}}. [/math]
The mechanics of division are a way to find a partial or complete factoring of a polynomial and do the possible canceling without any real thought.