How do you find the product #(3/4k+8)^2#?

2 Answers
Nov 5, 2017

Answer:

#(3/4k + 8)^2 = 9/36k^2 + 12k + 64#

Explanation:

#(x+y)^2 = x^2 + 2xy + y^2#

Nov 5, 2017

Answer:

#(9k^2)/16 +12k +64#

Explanation:

The mistake most people make is simply squaring both terms individually, so you to avoid this error, you should start by writing out the binomial like this:
#(3/4k +8)(3/4k+8)#

This is because it was squared; if it was cubed, you'd write it thrice, and so on.

This makes it easier to visualise what exactly you're doing, therefore you're less likely to make mistakes when solving it.

Now, I don't know if you've heard of FOIL method but I'll link a video here so you can understand it better before continuing with the question: enter link description here

So, using the FOIL method, we need to multiply:
the first terms of each binomial. i.e. #3/4k * 3/4k# which is #(9k^2)/16#
Then the outer terms, i.e. #3/4k*8# which gives #(24k)/4#
Then the inside terms, i.e #8*3/4k# which is #(24k)/4#
And finally the last terms, i,e, #8*8# which is #64#

Now you'll need to add these terms:
#(9k^2)/16 + (24k)/4 +(24k)/4+64#

[you can add the terms with the same variable; but remember: you can't add #k^2# and #k# because they have different powers.]

So,
#(9k^2)/16+ (48k)/4+64#

You can also now simplify the individual fractions, which is:
#(9k^2)/16 +12k +64#