A mother’s age is the product of the sum and difference of the ages of her two children. If x and y are the ag

Question:A mother’s age is the product of the sum and difference of the ages of her two children. If x and y are the ages of the children, express the mother’s age as a polynomial.

Please show how you get the answer please so I will understand.

Answers:
The previous answers are correct, but you need to continue the multiplication to get the polynomial.

(x-y)*(x+y)

= x(sqaured) + xy - xy - y(squared)

= x(sq) - y(sq)

{sorry I don't know how to type the exponents here}
Mothers age=M

X-childs age.
Y-Childs age.

(X+Y)(X-Y)=M

I suppose.
Ready? Let Mom be z. "Her age is" means
z=

Now, one child is x and the other is y.
the sum (means addition) of their ages is (x + y)
the difference (means subtraction) of their ages is (x-y)
What does product mean? Multiplication right?

so z=(x+y)(x-y)

Hope the explanation helps.
Okay, we can use the variable "z" for the mother's age. and "x" for child one and "y" for child two.

So, z=(x+y)(x-y)
Hope I helped!
The product is the answer to a multiplication sum, so if the

children are x and y years old, then the sum of their ages is

[x+y] years, and the difference of their ages is [x-y] years, so

Mother`s age is;

[x+y][x-y]

=x[x-y]+y[x-y]

=x^2-xy+xy-y^2

=x^2-y^2 years.

Hope this helps, Twiggy.
ma = (x+y) * (x-y)
multiply the brackets x(x-y) + y(x-y) = x^2 -xy +yx -y^2
ma = x^2 -xy +yx - Y^2 # -xy +yx = 0
ma = x^2 - y^2
age=[x+y] [x-y]

This article contents is post by this website user, EduQnA.com doesn't promise its accuracy.



More Questions & Answers...
  • Please help me with my Homework!?
  • History questin please help me?
  • If a and b are different positive integers and a2+b2<150, what is the greatest possible sum of a and b?
  • What r sum ideas for girls lockers?
  • I'm stuck on 2 math problems, can someone help me?
  • The federal government derives their power from.?
  • When analyzing a novel?
  • Hey does anybody know wat the circle of courage is?!?
  • Copyright 2006-2009 EduQnA.com All Rights Reserved.