Question

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

Answer 1

`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`