Factoring polynomial
- Thread starter lmf0610
- Start date
Hi Lisa!
Ok, first, I want you to know that it makes life just a bit easier if you actually write out your equation like this...
9x<sup>3</sup> + 39x<sup>2</sup> + 12 =
In case you don't know how to do this, all you have to do is write 9x then < sup> without the space and then write the number that you want to be up, in this case it's 3, and then end it by writing </ sup> without the space. This makes it look like this : 9x<sup>3</sup>.
Ok, first, I want you to know that it makes life just a bit easier if you actually write out your equation like this...
9x<sup>3</sup> + 39x<sup>2</sup> + 12 =
In case you don't know how to do this, all you have to do is write 9x then < sup> without the space and then write the number that you want to be up, in this case it's 3, and then end it by writing </ sup> without the space. This makes it look like this : 9x<sup>3</sup>.
That cubic has no rational zeros.
. . I'll bet there's a typo . . .
How about: \(\displaystyle \;9x^3\,+\,39x^2\,+\,12x\:=\:0\) ?
. . Factor: \(\displaystyle \;3x(x^2\,+\,13x\,+\,4)\:=\:0\)
. . Factor: \(\displaystyle \;3x(x\,+\,4)(3x\,+\,1)\:=\:0\)
Therefore: \(\displaystyle \;x\,=\,0,\,-4,\,-\frac{1}{3}\)
. . I'll bet there's a typo . . .
How about: \(\displaystyle \;9x^3\,+\,39x^2\,+\,12x\:=\:0\) ?
. . Factor: \(\displaystyle \;3x(x^2\,+\,13x\,+\,4)\:=\:0\)
. . Factor: \(\displaystyle \;3x(x\,+\,4)(3x\,+\,1)\:=\:0\)
Therefore: \(\displaystyle \;x\,=\,0,\,-4,\,-\frac{1}{3}\)