# Factor `20g^3-4g^2-25g+5`Factor by grouping, if possible, and check

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Factor `20g^3-4g^2-25g+5` :

The first two terms have a common factor of `4g^2` , while the last terms have a common factor of -5; factoring out the common terms yields:

`4g^2(5g-1)-5(5g-1)` . These terms have a common factor of (5g-1), so we factor this out to get:

`(5g-1)(4g^2-5)` . The last factor does not factor in the rationals, so we are done.

**The factored form of** `20g^3-4g^2-25g+5=(5g-1)(4g^2-5)`

** Note that multiplying the binomials yields `20g^3-4g^2-25g+5` **

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If you want the polynomial fully factored you will get:

`(5g-1)(2g+sqrt(5))(2g-sqrt(5))` -- multiplying these binomials will also get back to the original problem.

**Sources:**

`(20g^3-4g^2)(-25g+5) ` the first one has `4g^2` in common and the second one has `-5` in common

`4g^2(5g-1) -5(5g+5)` now set up the two numbers outside together:

`(5g-1) (4g^2-5)` and that's it's factored state, you can try to solve the parenthesis