Completely Factor 32x^5 + 243y^5 Using Synthetic Division | Step-by-Step Guide

[Plutonium88]

Completely factor 32x^5 + 243y^5

okay i used synthetic divison and i referenced a factorign calculator to get the answer

Answer: i used f(x) = (2x)^5 + (3y)^5

then f(-3y/2)= (2x)^5 + (3y)^5 = 0 (factor thereom)

using synthetic divison and factor (-3y/2)

i got.. 32x^4 -48x^3y + 72x^2y^2 -108xy^3 + 162y^4
The answer is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4)
i notice that the second piece (16x^4...)

the whole piece is actually half of what i long divided..so my problem is why am i not dividing to get that answer? and how can i put together what i knowÉ

 

[Mark44]

Completely factor 32x^5 + 243y^5

okay i used synthetic divison and i referenced a factorign calculator to get the answer

Answer:


i used f(x) = (2x)^5 + (3y)^5

then f(-3y/2)= (2x)^5 + (3y)^5 = 0 (factor thereom)

using synthetic divison and factor (-3y/2)

i got.. 32x^4 -48x^3y + 72x^2y^2 -108xy^3 + 162y^4

So what you have is
(2x)⁵ + (3y)⁵ = (x - (-3y/2))(32x⁴ -48x³y + 72x²y² -108xy³ + 162y⁴)
= (x + 3y/2)(32x⁴ -48x³y + 72x²y² -108xy³ + 162y⁴)

If you pull a factor of 2 out of the long factor, and use it to multiply the first factor, you get what you show below.

The answer is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4)



i notice that the second piece (16x^4...)

the whole piece is actually half of what i long divided..


so my problem is why am i not dividing to get that answer? and how can i put together what i knowÉ

 

[Plutonium88]

So what you have is
(2x)⁵ + (3y)⁵ = (x - (-3y/2))(32x⁴ -48x³y + 72x²y² -108xy³ + 162y⁴)
= (x + 3y/2)(32x⁴ -48x³y + 72x²y² -108xy³ + 162y⁴)

If you pull a factor of 2 out of the long factor, and use it to multiply the first factor, you get what you show below.

Just out of curiosity why am i allowed to divide a two out of the long factor, but i have to multiply it by the small factor? Like why is that allowed to happen, is there a rule of some sort... :O

 

[scurty]

Just out of curiosity why am i allowed to divide a two out of the long factor, but i have to multiply it by the small factor? Like why is that allowed to happen, is there a rule of some sort... :O

Maybe this will help clear the confusion up:

Consider
[itex]
\begin{eqnarray*}
(\frac{a}{2}+\frac{b}{2})(2a + 2b) &=& \frac{2}{2} (\frac{a}{2}+\frac{b}{2})(2a + 2b)\\
&=& 2 \cdot (\frac{a}{2}+\frac{b}{2}) \cdot \frac{(2a + 2b)}{2}\\
&=& (a+b)(a + b)
\end{eqnarray*}
[/itex]

Does that help clear up what was done above?

 

[Plutonium88]

Maybe this will help clear the confusion up:

Consider
[itex]
\begin{eqnarray*}
(\frac{a}{2}+\frac{b}{2})(2a + 2b) &=& \frac{2}{2} (\frac{a}{2}+\frac{b}{2})(2a + 2b)\\
&=& 2 \cdot (\frac{a}{2}+\frac{b}{2}) \cdot \frac{(2a + 2b)}{2}\\
&=& (a+b)(a + b)
\end{eqnarray*}
[/itex]

Does that help clear up what was done above?

So is this kind of a choice, of expanding the 2 within the first factor and using the 1/2 and targeting the second factor peicce with it.. argh I am just havin a rough time with this idea..

the onyl connection i can make (whichi 'm not sure is true) is it looks like you multiply the equation by (2/2) to satisfy both the top and bottom of the equation, and then the rules of math just do the rest... but that's just a guess..

Can u explain to me why you multiply the equation by 2/2

 

[scurty]

I was just illustrating why you can divide one term by 2 and multiply the other by 2. It's the same thing as multiplying by [itex]\frac{2}{2} = 1[/itex]. I just figured it was easier to see in a smaller example.

 

[Plutonium88]

i understand that connection and looking at the smaller example in terms of the math(that is very clear).

I think i was just thinking about it to hard, or perhaps not in the right manner. thanks a gain for your help man.