# Expanding and Expanding brackets with FRACTIONAL POWERS

• February 23rd 2013, 01:22 PM
gamei
Expanding and Expanding brackets with FRACTIONAL POWERS
Hey guys,
bit stuck on how to expand+simply this equation. could you please help me out and post how to do get to the simplified equation?

the equation being: Attachment 27204

Gamei
• February 23rd 2013, 01:51 PM
Re: Expanding and Expanding brackets with FRACTIONAL POWERS
You use the foil method as you would when multiplying polynomials of integer exponents. for a pair of binomials the foil method is (a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd (rearranging for like terms afterward)

so in this case we have
$(2x^\frac{1}{2}-3y^\frac{3}{2})(2x^\frac{1}{2} + 3y^\frac{3}{2})$
$(2x^\frac{1}{2})(2x^\frac{1}{2}) + (2x^\frac{1}{2})(3y^\frac{3}{2}) - (2x^\frac{1}{2})(3y^\frac{3}{2}) - (3y^\frac{3}{2})(3y^\frac{3}{2})$
$(2x^\frac{1}{2})(2x^\frac{1}{2}) - (3y^\frac{3}{2})(3y^\frac{3}{2})$
$4x - 9y^3$ by exponent rules. Recall $(a^b)(a^c) = a^{b+c}$
• February 23rd 2013, 01:55 PM
Plato
Re: Expanding and Expanding brackets with FRACTIONAL POWERS
Quote:

Originally Posted by gamei
Hey guys,
bit stuck on how to expand+simply this equation. could you please help me out and post how to do get to the simplified equation?
the equation being: Attachment 27204

Oh come on, the sum and difference of squares: $2x-9y^3$