can somone help me with these 2 problems
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
can somone help me with these 2 problems
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
I'm assuming the question is to solve these equations equal to zero?
For the first equation you have $\displaystyle 12x^3+28x^2-27x-63=0$ yes?
You can factorise this into $\displaystyle (2x-3)(6x^2+23x+21)$ and then solve this either by factorising, or using the quadratic formula.
For the second, you have $\displaystyle 216x^3-1000=0$ which you can reduce to $\displaystyle 27x^3-125=0$ or $\displaystyle 27x^3=125$. Then take the cube root of each side.
Hope this helps.
Why would you assume that? The post said "factor", why not factor?
Actually, it was $\displaystyle 216x^3+ 1000$ but it is still true thatFor the first equation you have $\displaystyle 12x^3+28x^2-27x-63=0$ yes?
You can factorise this into $\displaystyle (2x-3)(6x^2+23x+21)$ and then solve this either by factorising, or using the quadratic formula.
For the second, you have $\displaystyle 216x^3-1000=0$ which you can reduce to $\displaystyle 27x^3-125=0$ or $\displaystyle 27x^3=125$. Then take the cube root of each side.
Hope this helps.
$\displaystyle a^3+ b^3= (a+ b)(a^2- ab+ b^2)$
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