How do you factor #y^3 + 3y^2 -4y +12# by grouping?

Question

Isaiah Anthony · Accepted Answer

See explanation below

As per given question, the polynomial can't be factorized by grouping. However I assume some possible typo in question. Assuming 1) #y^3+3y^2+4y+12# #=y^2(y+3)+4(y+3)# #=(y+3)(y^2+4)# 2) #y^3+3y^2-4y-12# #=y^2(y+3)-4(y+3)# #=(y+3)(y^2-4)# #=(y+3)(y+2)(y-2)#

Ethan Adkins · Answer

undefined To factor the expression $y^3 + 3y^2 - 4y + 12$ by grouping, you can group the terms in pairs:

$y^3 + 3y^2$ - $4y + 12$

From the first two terms, you can factor out the common factor $y^2$, and from the last two terms, you can factor out the common factor $-4$:

$y^2(y + 3)$ - $4(y - 3)$

Now, you can see that both grouped terms have a common factor of $(y + 3)$, so you can factor that out:

$(y^2 - 4)(y + 3)$

Finally, you can factor the expression $y^2 - 4$ further as the difference of squares:

$(y - 2)(y + 2)(y + 3)$

So, the factored form of $y^3 + 3y^2 - 4y + 12$ by grouping is $(y - 2)(y + 2)(y + 3)$.