Asked by Anonymous
I need help with solving these problems. When I checked it it didn't equal to the original equation.
Factor each completely.
1. 4n^3+4n^2-16n-16
2. 9k^3-15k^2-3k+5
3. 60m^3+48m^2-20m-16
4. 12n^3+8n^2-3n-2
Factor each completely.
1. 4n^3+4n^2-16n-16
2. 9k^3-15k^2-3k+5
3. 60m^3+48m^2-20m-16
4. 12n^3+8n^2-3n-2
Answers
Answered by
Steve
Looks like an exercise set using factoring by grouping.
4n^3+4n^2-16n-16
4(n^3+n^2-4n-4)
4(n^3-4n + n^2-4)
4(n(n^2-4) + (n^2-4))
4(n+1)(n^2-4)
4(n+1)(n-2)(n+2)
9k^3-15k^2-3k+5
3k(k^2-1) - 5(k^2-1)
(3x-5)(k-1)(k+1)
60m^3+48m^2-20m-16
4(15m^3 + 12m^2 - 5m - 4)
4(5m(3m^2-1) + 4(3m^2-1))
4(5m+4)(3m^2-1)
4(5m+4)(√3m-1)(√3m+1)
12n^3+8n^2-3n-2
3n(4n^2-1) + 2(4n^2-1)
(3n+2)(2n-1)(2n+1)
4n^3+4n^2-16n-16
4(n^3+n^2-4n-4)
4(n^3-4n + n^2-4)
4(n(n^2-4) + (n^2-4))
4(n+1)(n^2-4)
4(n+1)(n-2)(n+2)
9k^3-15k^2-3k+5
3k(k^2-1) - 5(k^2-1)
(3x-5)(k-1)(k+1)
60m^3+48m^2-20m-16
4(15m^3 + 12m^2 - 5m - 4)
4(5m(3m^2-1) + 4(3m^2-1))
4(5m+4)(3m^2-1)
4(5m+4)(√3m-1)(√3m+1)
12n^3+8n^2-3n-2
3n(4n^2-1) + 2(4n^2-1)
(3n+2)(2n-1)(2n+1)
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