# Factor completely: y^4+3y^3-5y^2-15y.

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### 3 Answers

We have to factorize : y^4+3y^3-5y^2-15y

y^4+3y^3-5y^2-15y

=> y(y^3 + 3y^2 - 5y - 15)

=> y(y^2(y + 3) - 5(y + 3))

=> y(y^2 - 5)(y + 3)

**The required factorization is y(y^2 - 5)(y + 3)**

The expression y^4+3y^3-5y^2-15y has to be factored. This can be done in the following way:

y^4+3y^3-5y^2-15y

y is common to all the terms

= y(y^3 + 3y^2 - 5y - 15)

In the terms y^3 + 3y^2, y^2 is common and in 5y + 15, 5 is common

= y(y^2(y + 3) - 5(y+3))

= y*(y^2 - 5)(y+3)

The factored form of y^4+3y^3-5y^2-15y is y*(y^2 - 5)(y+3)

We'll group the first 2 terms and the last 2 terms:

(y^4+3y^3) + (- 5y^2 - 15y)

We'll factorize by y^3 the first group and by -5y the second group:

y^3(y + 3) - 5y(y + 3)

We'll factorize by y(y + 3):

y(y+3)(y^2 - 5)

So, the result of factorization is:

(y^4+3y^3- 5y^2 - 15y) = y(y+3)(y^2 - 5)