# How do you find all rational roots for #4y^5 + 8y^4 - 29y^3 - 42y^2 + 45y + 54 = 0#?

##### 1 Answer

#### Answer:

Zeros:

#### Explanation:

#f(y) = 4y^5+8y^4-29y^3-42y^2+45y+54#

By the rational root theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-2, +-9/4, +-3, +-9/2, +-6, +-27/4, +-9, +-27/2, +-18, +-27, +-54#

That's rather a lot of possibilities to try, but first note that:

#f(-1) = -4+8+29-42-45+54 = 0#

So

#4y^5+8y^4-29y^3-42y^2+45y+54#

#=(y+1)(4y^4+4y^3-33y^2-9y+54)#

Sticking with simpler calculations, let's try substituting

#4y^4+4y^3-33y^2-9y+54#

#=4(16)+4(8)-33(4)-9(2)+54#

#=64+32-132-18+54 = 0#

So

#4y^4+4y^3-33y^2-9y+54 = (y-2)(4y^3+12y^2-9y-27)#

In the remaining cubic, the ratio of the first and second terms is the same as the ratio between the third and fourth terms. Hence it will factor by grouping:

#4y^3+12y^2-9y-27#

#=(4y^3+12y^2)-(9y+27)#

#=4y^2(y+3)-9(y+3)#

#=(4y^2-9)(y+3)#

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

#=(2y-3)(2y+3)(y+3)#

Hence three more zeros: