Question

How do you solve `3r ^ { 2} - 8r - 16= 0`?

Answer 1

`r=-4, r=-4/3`

Explanation:

This is a factoring problem.

You start by multiplying `3` by `-16`, which gives you `-48`.

Next, try to find two numbers that multiply to be `-48` and add to be `-8`.

Those numbers are `4` and `-12`
(`4-12=-8`) and (`4*-12=-48`)

Then, split the middle term in the equation, using those numbers.
`3r^2+4r-12r-16=0`

Look at the first two terms, and pull out what you can.
`r(3r+4)`
Then look at the next two terms, and do the same.
`4(3r+4)`

The full equation is now:
`r(3r+4)+4(3r+4)=0`

Because `3r+4` is common in both terms, you can "pull it out".
`(3r+4)(r+4)=0`

Then, you set each factor equal to zero and solve.
`3r+4=0`
`3r+4-4=0-4`
`3r=-4`
`r=-4/3`

`r+4=0`
`r+4-4=0-4`
`r=-4`

Answer 2

`r` equals `-4/3` and `4`

Explanation:

First let's see what we can do about the coefficients. Are there any common factors between `3, 8, 16`? No there aren't since `3` is prime and the other two aren't multiples of `3`.

Now we need to factor. When the coefficient is not equal to `1` we have to use a method that isn't as straight forward. Some people like to use the box method but I prefer to use an alternative method called factor by grouping

First find the factor:

`3r^2color(pink)(-8)r-16`

`color(white)(.)+color(white)(.)8`
`color(white)(.)xxcolor(white)(.)48`
....................
`color(white)(.) 1 color(white)(..-) 48`
`color(white)(.) 2 color(white)(..-) 24`
`color(white)(.) 3 color(white)(..-) 16`
`color(red)(color(white)(.) 4 color(white)(..) -12` `= color(pink)(-8)`

Now we fill the blanks with the factors. It doesn't matter which one goes where, just make sure to multiply them by `r`:

`( color(blue)(3r^2) + color(white)(-12r) ) + ( color(white)(4r) - color(blue)(16) )`

`( color(blue)(3r^2) + color(red)(-12r) ) + ( color(red)(4r) - color(blue)(16) )`

Factor out the greatest common factor:

`3r(r-4)+4(r-4)`

The parentheses have the same value, so we're good. Factor `(r-4)` from the expression:

`(r-4) (3r+4)`

Now we have our final factored form. but we still haven't solved the problem. Set each factor equal to `0` and solve:

`* * * * * * * * * * * * * * * * * * * * * * * * * * *`

`r-4 = 0`

`r=4`

`* * * * * * * * * * * * * * * * * * * * * * * * * * *`

`3r+4 = 0`

`3r = -4`

`r=-4/3`

`* * * * * * * * * * * * * * * * * * * * * * * * * * *`

So, `r` equals `-4/3` and `4`