My first question is f(x)=9x^4-27
*must factor out completly (don't forget to include complex roots)
---- i don'yt know where to start. i can't use the quadratic---since it is raised to the 4th how would i factor out? would it be
9x^4+27x^2-27-27 but then i have a x^2 in my equation.
my second question is
Solve 3x^2+7x=-45
---- if i use the quatric in this equation i get a decimal under my root. So i'm stuck after i equal it to zero.
3x^2+7x+45=0
i'm having trouble factoring this out. i need 2 numbers that when mult equal 45 and when added/subtracted equal 7? but both of my signs need to be (+) positive?
Help please
I would use " ". Here a= 3x, obviously. 27 doesn't look like "b^2" but .
.
We can do the same thing with 3x^2- 3\sqrt{3}: it is of the form with and . .
only requires that we put in [itex]-1= i^2[/itex]. [tex]3x^2+ 3\sqrt{3}= 3(x+ i\sqrt[4]{3})(x- i\sqrt[4]{3}).
Putting those together, = 9(x+ \sqrt[4]{3})(x- \sqrt[4]{3})(x+ i\sqrt[4]{3})(x- i\sqrt[4]{3})
Having done that, it occurs to me that the easiest way to factor a polynomial is to set it equal to 0 and solve the equation! If , then has roots , , , and which tells us that and so as above.
quartic? This is quadratic and the quadratic formula works nicely: has roots . 491 happens to be prime so that square root cannot be reduced but we can writemy second question is
Solve 3x^2+7x=-45
---- if i use the quatric in this equation i get a decimal under my root. So i'm stuck after i equal it to zero.
3x^2+7x+45=0
i'm having trouble factoring this out. i need 2 numbers that when mult equal 45 and when added/subtracted equal 7? but both of my signs need to be (+) positive?
Help please