Question

How do you factor the trinomial `32z^5-20z^4-12z^3`?

Answer 1

`32z^5 - 20z^4 - 12z^3 = 4z^3(z-1)(8z+3)`

Explanation:

Noting that each term has a factor of `z^3` and that the greatest common divisor of `32`, `20`, and `12` is `4`, we may first factor those out to obtain

`32z^5 - 20z^4 - 12z^3 = 4z^3(8z^2 - 5z - 3)`

Next, we can check to see how to factor `8z^2 - 5z - 3`.
As `z = 1 => 8z^2 - 5z - 3 = 0` we know that `(z - 1)` will be a factor. So all we need is to figure out what `a` and `b` gives us
`(z-1)(az+b) = 8z^2 - 5z - 3`.

Because `z * az = 8z^2` we can tell that `a = 8`.
Because `-1*b = -3` we can tell that `b = 3`.

So `(z-1)(8z+3) = 8z^2 - 5z - 3`.

Substituting this back in gives our final result:

`32z^5 - 20z^4 - 12z^3 = 4z^3(z-1)(8z+3)`

Answer 2

`4z^3(8z+3)(z-1)`

Explanation:

`color(blue)(Step 1)`
Are we able to turn this partly into a quadratic form?
Yes if we factor out as many `z's` as possible. We can also factor out 4.

Write as: `4z^3(8z^2 -5z-3)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)(Step 2)`

Consider just the brackets
We know that the product of 2 numbers give us the -3.
We also know that the product of 2 numbers give us the 8 in `color(white)(.)8z^2`

Factors of 8 `->{1,8} , {2,4}`
Factors of 3 `->{1,3}`

As there are only 2 factors of three we are safe in writing:
`4z( ? .. 1)(?.. 3)`

Now consider the factors of 8
Suppose we used {2,4}

`color(brown)(Try color(white)(.) 1)`

`(2...1)(4...3) -> (2 times 3) +-(1 times 4) != 5 color(white)(..) color(red)("Fail!")`

`color(brown)(Try color(white)(.) 2)`

`(2...3)(4...1) -> (2 times 4) -(3 times 1) = 5 color(white)(..) color(red)("Fail!")` we need -5

`color(brown)(Try color(white)(.) 3)`

`(8z +3 )(z -1 ) = 8z^2 +3z-8z-3color(white)(..) color(green)("Works!")`

`color(blue)("Putting it all together")`

`color(green)(4z^3(8z+3)(z-1)) =4z^3(8z^2-5z-3) =32z^5 -20z^4-12z^3`