How do you factor #y^4-14y^3+49y^2#?

1 Answer
Sep 18, 2016

Answer:

#y^2(y-7)#

Explanation:

Take out a common factor of #y^2# first.

#y^2(y^2-14y color(red)(+49)) " "rarr color(red)(+49)# a clue!

You should recognize #49# as being a perfect square!

In #(x+y)^2 = x^2 +2xy +y^2 = ax^2+bx + c# there is always a special relationship between the coefficient of the second term (b) and the third term (c)

#(bdiv 2) " and then squared gives " c#.

If this relationship exists in the trinomial you will know you have a square of a binomial.

Let's look closer..... #(14div2)^2 = 7^2=49!!#

#:. y^2color(red)((y^2-14y+49)) = y^2color(red)((y-7)^2)#

If you missed that, you can try to find factors of 49 which add to 14. This will give you factors of 7 and 7 anyway:

#y^2(y-7)(y-7)#

=#y^2(y-7)^2#