Faster Polynomial Multiplication

In summary, the conversation discusses the correct way to solve equations involving factors that are sums of powers. The correct formula to use is (x + y)(x - y) = x^2 - y^2. The conversation also includes an example where the formula was applied incorrectly and the correct way to solve the equation.
  • #1
Kys91
12
0
I can solve:

(2x+3y-4z)(2x-3y+4z) = [2x +(3y-4z)] [2x -(3y-4z)] = (2x)^2 -(3y-4z)^2 = (2x)^2 - (9y^2 - 24yz - 16z) = 4x^2 - 9y^2 + 24yz + 16z which is fine, but if I try to solve:

(m^2-m-1)(m^2+m-1) = [m^2 -(m+1)][m^2+(m-1) = (m^2)^ 2 - (m-1)^2 = (m^2)^2 - (m^2 -2m +1) = m^4 - m^2 +2m -1 which is not, and I also tried doing:

[(m^2 - m)-1] [(m^2+m)-1] = (m^4 - m^2) + 1 = this latest one is near, but anyway it should be m^4-3m^2 + 1

Doing normal multiplication I get: m^4 +m^3 -m^2 -m^3 -m^2+ m-m^2-m+ 1 = m^4-3m^2 + 1

So why is it that I have -m^2 when is it really -3m^2?

Thanks
 
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  • #2
Since you have factors which are sums of powers of m, use a lattice system to help keep like terms easily found.
 
  • #3
You are using the equality: (x + y)(x - y) = x2 - y2, right?

Kys91 said:
...

(m^2-m-1)(m^2+m-1) = [m^2 -(m+1)][m^2+(m-1) = (m^2)^ 2 - (m-1)^2 = (m^2)^2 - (m^2 -2m +1) = m^4 - m^2 +2m -1 which is not, and I also tried doing:

Notice the bolded part, that's where you went wrong. Since m + 1 does not equal m - 1, so you cannot apply the equality here. Remember that:

(x + y)(x - y) = x2 - y2

[(m^2 - m)-1] [(m^2+m)-1] = (m^4 - m^2) + 1 = this latest one is near, but anyway it should be m^4-3m^2 + 1

This is totally wrong, since m2 - m is not the same as m2 + m.

And you are having 2 minus signs here (the red part). It should be one plus, and one minus instead. Please stick to the formula! It's NOT: (x - y)(x - y) = x2 + y2! This is nowhere near correct!

Instead, what we should use here is:

(x + y)(x - y) = x2 - y2

-----------------------

So, back to your problem:

(m2 - m - 1)(m2 + m - 1)

Notice the 2 terms -m, and +m, they are of different signs. Let's see if you can find any way to apply the equality (x + y)(x - y) = x2 - y2 here. :)
 
Last edited:
  • #4
Kys91 said:
I can solve:

(2x+3y-4z)(2x-3y+4z) = [2x +(3y-4z)] [2x -(3y-4z)] = (2x)^2 -(3y-4z)^2 = (2x)^2 - (9y^2 - 24yz - 16z) = 4x^2 - 9y^2 + 24yz + 16z which is fine, but if I try to solve:

(m^2-m-1)(m^2+m-1) = [m^2 -(m+1)][m^2+(m-1) = (m^2)^ 2 - (m-1)^2 = (m^2)^2 - (m^2 -2m +1) = m^4 - m^2 +2m -1 which is not, and I also tried doing:

[(m^2 - m)-1] [(m^2+m)-1] = (m^4 - m^2) + 1 = this latest one is near, but anyway it should be m^4-3m^2 + 1

Doing normal multiplication I get: m^4 +m^3 -m^2 -m^3 -m^2+ m-m^2-m+ 1 = m^4-3m^2 + 1

So why is it that I have -m^2 when is it really -3m^2?

Thanks
Because you did it wrong, of course!:wink:
You have [itex][(m^2- m)-1][(m^2+m)-1][/itex] and did the first part as a "sum and difference product": [itex](m^2- m)(m^2+ m)= (m^2)^2- m^2= m^4- m^2[/itex] which is correct. And, of course, (1)(1)= 1. But you forgot the "middle terms" (the "O" and "I" of "FOIL").
[tex][(m^2- m)- 1][(m^2+m)- 1]= (m^2- m)(m^2+m)+ (m^2-m)(-1)+ (-1)(m^2+m)+(-1)(-1)[/tex]
[tex]= (m^4- m^2)- (m^2- m)- (m^2+ m)+ 1= m^4- 3m^2+ 1[/tex]
(The "m" and "-m" terms cancel.)
 
  • #5
Alright guys, thanks. :smile: Studying yourself + this forum is a lethal formula for learning about a lot of things.
 

1. How does faster polynomial multiplication work?

Faster polynomial multiplication uses algorithms and techniques to reduce the number of operations needed to multiply two polynomials. This results in a faster and more efficient way of computing polynomial products.

2. What are the benefits of faster polynomial multiplication?

The main benefit of faster polynomial multiplication is that it significantly reduces the time and resources needed to perform polynomial multiplication. This is especially useful for large or complex polynomials that would otherwise take a long time to compute.

3. Can faster polynomial multiplication be applied to all types of polynomials?

Yes, faster polynomial multiplication can be applied to any type of polynomial, including monomials, binomials, and polynomials with higher degrees. However, the level of efficiency may vary depending on the specific algorithm used.

4. Are there any disadvantages to using faster polynomial multiplication?

One potential disadvantage of faster polynomial multiplication is that the algorithms used may be complex and difficult to understand, making it challenging for beginners to apply. Additionally, some algorithms may only be efficient for certain types of polynomials, so it is important to choose the right algorithm for the specific problem.

5. How does faster polynomial multiplication compare to traditional polynomial multiplication?

In general, faster polynomial multiplication is significantly faster and more efficient than traditional polynomial multiplication. This is because it uses advanced algorithms and techniques to reduce the number of operations needed. However, for simpler polynomials, the difference in speed may not be as significant.

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