MHB How to Distribute in an Algebra Problem

  • Thread starter Thread starter tmt1
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
To simplify the expression 2(x^2+y^2)(2x+2yy'), distribute (x^2+y^2) across the terms in parentheses. This results in 4x(x^2+y^2) + 4y(x^2+y^2)y'. The key is recognizing that each term in the parentheses must be multiplied by the factor outside. This process clarifies the transition between the two forms of the equation. Understanding distribution is essential for solving similar algebraic problems.
tmt1
Messages
230
Reaction score
0
Hi working on an algebra/calculus problem

How do I get from

2 (x^2+y^2) (2x+2yy')

to 4x (x^2+y^2) + 4y(x^2+y^2) y'

It is very confusing.

Tim
 
Last edited:
Mathematics news on Phys.org
tmt said:
Hi working on an algebra/calculus problem

How do I get from

2 (x2+y2) (2x+2yy')

to 4x (x2+y2) + 4y(x2+y2) y'

It is very confusing.

Tim

Did you mean, how do you get from $2(x^{2}+y^{2})(2x+2yy')$ to $4x(x^{2}+y^{2})+4y(x^{2}+y^{2})y'$? Because, if so, that's just distributing the $(x^{2}+y^{2})$ times the other stuff in parentheses.
 
Ackbach said:
Did you mean, how do you get from $2(x^{2}+y^{2})(2x+2yy')$ to $4x(x^{2}+y^{2})+4y(x^{2}+y^{2})y'$? Because, if so, that's just distributing the $(x^{2}+y^{2})$ times the other stuff in parentheses.
Yes I did,

Thank you!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top