# use factoring and the zero product property to solve the problem9r^2-30r+21=-4

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`9r^2-30r+21 = -4`

To solve this kind of problem we have to set one side of the equation to 0.

`9r^2-30r+21+4 = 0`

`9r^2-30r+25 = 0`

When the equation set to zero in one side we have to do the other side to solve it for the unknown.

First we will consider a general quadratic equation which is similar to this.

`ar^2+br+c = 0`

According to our question `a = 9` , `b = -30` and `c = 25`

Then consider the product of `a` and `c` which is `(axxc)`

`axxc = 9xx25 = 225`

Now factor the product of `a` and `c` in such a way that the individual factors has a sum equal to `b` .

`225 = 25xx9 rarr (25+9 = 34!=b)`

`225 = 15xx15 rarr (15+15 = 30!=b)`

`225 = -15xx-15 rarr (-15+(-15) = -30 = b)`

Now replace the b component as a addition of the factors you found.

`9r^2-30r+25 = 0`

`9r^2-15r-15r+25 = 0`

Now simplify the first two terms and last two terms separately getting common terms out. In this process make sure that you will get same answer in both brackets.

`3r(3r-5)-5(3r-5) = 0`

`(3r-5)(3r-5) = 0`

For this to happen `(3r-5) = 0` . So `r = 5/3`

*So the answer is r = 5/3.*

*Here Both factors comes as 3r-5. But it doesn't happen always. You will get two factors most of the times.*