Question

Factor the expression: -\lambda ^{3}+4\lambda ^{2}+4\lambda -16

Answer 1

Answer:

$-(\lambda -4)(\lambda -2)(\lambda +2)$

Step-by-step explanation:

What we want to fator is:

$-\lambda^3 + 4 \lambda^2 + 4\lambda - 16$

There is no common factor, but let's factor it by grouping. The first two addends can be factor as follows:

$-\lambda^3 + 4 \lambda^2 = \lambda^2(-\lambda + 4) = -\lambda^2(\lambda - 4)$

the second addends can be factor as well:

$4\lambda - 16 = 4(\lambda- 4)$.

Then our original expression can be rewritten like

$-\lambda^3 + 4 \lambda^2 + 4\lambda -16=\lambda^2(\lambda - 4) + 4(\lambda - 4)$

And here the $(\lambda-4)$ is the common factor!

$-\lambda^2(\lambda - 4) + 4(\lambda - 4) = (\lambda - 4)(-\lambda^2 + 4)$

Finally, we can factor the quadratic expression as a difference of squares $-\lambda^2 + 4 = 4 - \lambda^2 = (2+\lambda)(2-\lambda)$

Ant we get

$(\lambda - 4)(-\lambda^2 + 4)= (\lambda - 4)(\lambda + 2)(2-\lambda)$

now, we can extract the negative sign from $(2-\lambda)$, and we get

$-(\lambda -4)(\lambda -2)(\lambda +2)$.