1. ## Factor and Solve

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?

2. Originally Posted by b_austintx
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?

The first one is repeated use of the Difference of Two Squares.

$\displaystyle 9x^4 - 27 = 9(x^4 - 3)$

$\displaystyle = 9\left[(x^2)^2 - \left(3^{\frac{1}{2}}\right)^2\right]$

$\displaystyle = 9\left(x^2 - 3^{\frac{1}{2}}\right)\left(x^2 + 3^{\frac{1}{2}}\right)$

$\displaystyle = 9\left[x^2 - \left(3^{\frac{1}{4}}\right)^2\right]\left[x^2 + \left(3^{\frac{1}{4}}\right)^2\right]$

$\displaystyle = 9\left[x + 3^{\frac{1}{4}}\right]\left[x - 3^{\frac{1}{4}}\right]\left[x^2 - \left(3^{\frac{1}{4}}i\right)^2\right]$

$\displaystyle = 9\left[x + 3^{\frac{1}{4}}\right]\left[x - 3^{\frac{1}{4}}\right]\left[x + 3^{\frac{1}{4}}i\right]\left[x - 3^{\frac{1}{4}}i\right]$.

3. Originally Posted by b_austintx
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.
I would use "$\displaystyle a^2- b^2= (a+ b)(a- b)$". Here a= 3x, obviously. 27 doesn't look like "b^2" but $\displaystyle 27= (\sqrt{27})^2= (3\sqrt{3})^2$.

$\displaystyle 9x^4- 27= (3x^2- 3\sqrt{3})(3x^2+ 3\sqrt{3})$.

We can do the same thing with 3x^2- 3\sqrt{3}: it is of the form $\displaystyle a^2- b^2$ with $\displaystyle a= x\sqrt{3}$ and $\displaystyle b= \sqrt{3}\sqrt[4]{3}$. $\displaystyle 3x^2- 3\sqrt{3}= (x\sqrt{3}+ \sqrt{3}\sqrt[4]{3})(x\sqrt{3}- \sqrt{3}\sqrt[4]{3})= 3(x+ \sqrt[4]{3})(x- \sqrt[4]{3})$.

$\displaystyle 3x^2+ 3\sqrt{3}= 3x^2- (-1)(3)\sqrt{3}$ only requires that we put in $-1= i^2$. [tex]3x^2+ 3\sqrt{3}= 3(x+ i\sqrt[4]{3})(x- i\sqrt[4]{3}).

Putting those together, $\displaystyle 9x^4- 27$= 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 $\displaystyle 9x^4- 27= 9(x^4- 3)= 0$, then $\displaystyle x^4= 3$ has roots $\displaystyle \sqrt[4]{3}$, $\displaystyle -\sqrt[4]{3}$, $\displaystyle i\sqrt[4]{3}$, and $\displaystyle -i\sqrt[4]{3}$ which tells us that $\displaystyle x^4- 3= (x- \sqrt[4]{3})(x+ \sqrt[4]{3})(x- i\sqrt[4]{3})(x+ i\sqrt[4]{3})$ and so $\displaystyle 9x^4- 27= 9(x- \sqrt[4]{3})(x+ \sqrt[4]{3})(x- i\sqrt[4]{3})(x+ i\sqrt[4]{3})$ as above.
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?

quartic? This is quadratic and the quadratic formula works nicely: $\displaystyle 3x^2+ 7x+ 45= 0$ has roots $\displaystyle x= \frac{-7\pm\sqrt{49-540}}{6}= \frac{-7\pm\sqrt{-491}}{6}= \frac{-7\pm i\sqrt{491}}{6}$. 491 happens to be prime so that square root cannot be reduced but we can write $\displaystyle 3x^3+ 7x+ 45= 3(x+\frac{7}{6}- i\frac{\sqrt{491}}{6})(x+\frac{7}{6}+ i\frac{\sqrt{491}}{6})$