Question

How do you factor the trinomial `7y^2- 50y + 7`?

Answer 1

`7y^2-50y+7 = (7y-1)(y-7)`

Explanation:

Let `f(y) = 7y^2-50y+7`

By the rational root theorem, any rational zeros of `f(y)` must be expressible in the form `p/q` for some integers `p` and `q` where `p` is a divisor of the constant term `7` and `q` is a divisor of the coefficient `7` of the leading term.

That means that the only possible rational zeros are:

`+-1/7, +-1, +-7`

In addition, since the coefficients are symmetric, if `y=r` is a zero then so is `y=1/r`.

Also we find that `f(-y) = 7y^2+50y+7` has no changes of sign in its coefficients, so `f(y)` has no negative zeros.

Also the sum of the coefficients is non-zero. That is `7-50+7 = -36 != 0`, so `y=1` is not a zero.

Hence if the zeros are rational, then they are `1/7` and `7`.

Try:

`(7y-1)(y-7) = 7y^2-50y+7`

Yes!

Answer 2

Use an AC Method to find:

`7y^2-50y+7=(7y-1)(y-7)`

Explanation:

Given `7y^2-50y+7` find a pair of factors of `AC=7*7=49` with sum `B=50`.

The pair `49, 1` works:

`49xx1 = 49`

`49+1 = 50`

Use this pair to split the middle term and factor by grouping:

`7y^2-50y+7`

`=7y^2-49y-y+7`

`=(7y^2-49y)-(y-7)`

`=7y(y-7)-1(y-7)`

`=(7y-1)(y-7)`