How do you solve 3r2−8r−16=0?

2 Answers
Elle
Jan 18, 2018

r=−4,r=−43

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.
3r2+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=−43

r+4=0
r+4−4=0−4
r=−4

Rachel
Jan 18, 2018

r equals −43 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:

3r2−8r−16

.+.8
.×.48
....................
.1..−48
.2..−24
.3..−16
.4..−12 =−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:

(3r2+−12r)+(4r−16)

(3r2+−12r)+(4r−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=−43

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

So, r equals −43 and 4

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