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

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### 3 Answers

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

First we will group the first two terms and the last two terms together:

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

Now we will factor y^3 from the first two terms:

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

Now we will factor 5y from the last two terms:

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

Now we notice that both terms have ( y+ 3) , then we will factor ( x+ 3) again:

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

Now we will factor y from (y^3 - 5y)

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

Now we can factor (y^2-5) at follows:

(y^2-5) = ( y-sqrt5)(y+sqrt5)

Then the last form is

**(y^4+3y^3-5y^2-15y = y*(y+3)*(y-sqrt5)(y+sqrt5) **

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)**

To factor y^4+3y^3-5y^2-15y.

Since y is a factor of each term, y is a factor.

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

We take the expression in the bracket for factorisation:

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

We take the factor y^2-5 which is equal to (y-sqrt5)(y+sqrt5).

Therefore , y^4+3y^3-5y^2-15y = y(y+3)(y-sqrt5)(y+sqrt5) which gives all the factors.