How do you simplify #(sqrtc + d)(3 +sqrt5)#?

1 Answer
Meave60
Apr 8, 2017

#(sqrtc+d)(3+sqrt5)=color(blue)(3sqrtc+sqrt5sqrtc+3d+dsqrt5#

Explanation:

Simplify:

#(sqrtc+d)(3+sqrt5)#

Use the FOIL method, which gives the order in which the terms of two binomials are multiplied: First, Outer, Inner, Last

http://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method.html

#(sqrtc+d)(3+sqrt5)=(sqrtc*3)+(sqrtc*sqrt5)+(d*3)+(d*sqrt5)#

#(sqrtc+d)(3+sqrt5)=3sqrtc+sqrt5sqrtc+3d+dsqrt5#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
944 views around the world
You can reuse this answer
Creative Commons License