How do you factor y^3-8y+2y^2-16y3−8y+2y2−16?

2 Answers
Don't Memorise
Jun 21, 2015

=color(red)((y+2) (y^2-8) =(y+2)(y2−8)

Explanation:

y^3-8y+2y^2-16y3−8y+2y2−16

Factorising by grouping:
Forming two groups of two terms:

color(red)(y^3-8y)+color(blue)(2y^2-16y3−8y+2y2−16
=y (y^2-8) + 2(y^2-8)=y(y2−8)+2(y2−8)
=color(red)((y+2) (y^2-8) =(y+2)(y2−8)

George C.
Jun 21, 2015

y^3-8y+2y^2-16y3−8y+2y2−16

=(y+2)(y^2-8)=(y+2)(y2−8)

=(y+2)(y-2sqrt(2))(y+2sqrt(2))=(y+2)(y−2√2)(y+2√2)

Explanation:

First factor by grouping

f(y) = y^3-8y+2y^2-16f(y)=y3−8y+2y2−16

=(y^3-8y)+(2y^2-16)=(y3−8y)+(2y2−16)

=y(y^2-8)+2(y^2-8)=y(y2−8)+2(y2−8)

=(y+2)(y^2-8)=(y+2)(y2−8)

The remaining quadratic term can be factorized by treating it as a difference of squares with irrational coefficients:

f(y) =(y+2)(y^2-(sqrt(8))^2)f(y)=(y+2)(y2−(√8)2)

=(y+2)(y-sqrt(8))(y+sqrt(8))=(y+2)(y−√8)(y+√8)

...using the difference of squares identity:

a^2-b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)

Finally note that sqrt(8) = sqrt(2^2*2) = 2sqrt(2)√8=√22⋅2=2√2

f(y)=(y+2)(y-2sqrt(2))(y+2sqrt(2))f(y)=(y+2)(y−2√2)(y+2√2)

Related questions
See all questions in Factoring Completely
Impact of this question
1834 views around the world
You can reuse this answer
Creative Commons License