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

  • Thread starter Plutonium88
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In summary, the equation is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4) and the answer is (2x+3y)(16x^4 -24x^3y + 36x^2y^2 - 54xy^3 + 81y^4).
  • #1
Plutonium88
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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
 
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  • #2
Plutonium88 said:
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.

Plutonium88 said:
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
 
  • #3
Mark44 said:
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
 
  • #4
Plutonium88 said:
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?
 
  • #5
scurty said:
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
 
  • #6
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.
 
  • #7
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.
 

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

What is synthetic division?

Synthetic division is a method used to divide a polynomial by a linear factor. It is often used to solve equations or find the roots of a polynomial.

Why would I want to use synthetic division?

Synthetic division can make dividing polynomials quicker and easier, especially when the divisor is a linear factor.

What is the step-by-step process for using synthetic division?

The first step is to set up the problem in the correct format, with the divisor on the outside and the coefficients of the dividend on the inside. Then, follow the steps of synthetic division, including bringing down the first coefficient, multiplying it by the divisor, adding it to the next coefficient, and repeating until all coefficients have been used. Finally, interpret the results to find the quotient and remainder.

Can synthetic division be used for any polynomial?

No, synthetic division can only be used when dividing by a linear factor. If the divisor is not linear, traditional long division must be used.

How can I check my answer after using synthetic division?

You can check your answer by multiplying the quotient by the divisor and adding the remainder. The result should be equal to the original dividend. Additionally, you can plug the roots (found in the quotient) back into the original polynomial to see if they result in a zero value.

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