can somone help me with these 2 problems (Crying)

1st problem 12x^3+28x^2-27x-63

2nd problem 216x^3+1000

if you could just show me step by step how its solved so i can figure out the other few i gota do and not use confusing math terms

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- March 23rd 2011, 01:48 PMshadower4factor each of the following completely
can somone help me with these 2 problems (Crying)

1st problem 12x^3+28x^2-27x-63

2nd problem 216x^3+1000

if you could just show me step by step how its solved so i can figure out the other few i gota do and not use confusing math terms - March 23rd 2011, 01:57 PMAlso sprach Zarathustra
Hints:

12x^3+28x^2-27x-63=4x^2(3x+7)-9(3x+7)

216x^3+1000=(6x)^3 + 10^3 - March 23rd 2011, 01:58 PMworc3247
I'm assuming the question is to solve these equations equal to zero?

For the first equation you have yes?

You can factorise this into and then solve this either by factorising, or using the quadratic formula.

For the second, you have which you can reduce to or . Then take the cube root of each side.

Hope this helps. - March 23rd 2011, 02:14 PMHallsofIvy
Why would you assume that? The post said "factor", why not factor?

Quote:

For the first equation you have yes?

You can factorise this into and then solve this either by factorising, or using the quadratic formula.

For the second, you have which you can reduce to or . Then take the cube root of each side.

Hope this helps.

- March 23rd 2011, 02:32 PMworc3247
My bad, I didn't read the title.

- March 24th 2011, 12:56 PMZap
Sometimes you can use the regroup and factor method on these 3rd order equations:

grouping the left 2 members then the right 2 members: (12x^3 + 28x^2) - (27x + 63)

factoring 4x^2 out of the left side and 9 out of the right side: 4x^2(3x +7) - 9(3x+7)

factoring (3x+7) from both sides: (3x+7)(4x^2-9)

factoring the quadratic is like when you factor (a^2-b^2) so you have: (3x+7)(2x+3)(2x-3)

and it looks like HallsofIvy gave you the formula for the x^3+y^3 one.. hope this is a bit clearer on the mechanics... grin