How do you multiply #(8x^2+3)(8x^2-3)#?

1 Answer
Aug 4, 2016

Answer:

#64x^4-9#

Explanation:

#color(blue)("Method 1")#

#color(red)((8x^2+3)color(purple)((8x^2-3)))#

Multiply everything inside one bracket by everything inside the other.

#color(purple)(color(red)(8x^2)(8x^2-3)" "color(red)(+3)(8x^2-3))#

#64x^4-24x^2 " "+24x^2-9#

#64x^4 + 0 -9#

#64x^4-9#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Method 2")#
The following is a general equation that is worth committing to memory:

Consider the general case of: #a^2+b^2=(a+b)(a-b)#

Think of #(8x^2+3)(8x^2-3)# as of form #(a+b)(a-b)#

This gives:#" "[(8x^2)^2-3^3] = 64x^4-9 larr" a 1 line solution"#

'..................................................................................
#color(brown)("If this is still not clear then let "u=8x^2" giving:")#

#(u+3)(u-3)=(u^2-3^2)#

But #u=8x^2# giving

#[(8x^2)^2-3^2]#

#=64x^4-9#