How do you multiply #(4n - 1) ( 3n + 4)#?

2 Answers
Sep 24, 2016

#12n^2+13n-4#

Explanation:

start off by writing out
#(4n-1)(3n+4)#

multiply by distributing
#(4n)(3n)=12n²#
#(4n)(4)=16n#

so far you have
#12n² +16n#

now distribute again but with the -1
#(-1)(3n)=-3n#
#(-1)(4)=-4#

you now have #12n^2 + 16n-3n-4#

subtract like terms
#(16n-3n)=13n#

put it together and you have:
#12n^2+13n-4#

Sep 24, 2016

Use FOIL to find:

#(4n-1)(3n+4) = 12n^2+13n-4#

Explanation:

If you find it helpful, you can use the FOIL ("First", "Outside", "Inside", "Last") mnemonic to help to combine all the required terms:

#(4n-1)(3n+4) = overbrace((4n*3n))^"First"+overbrace((4n*4))^"Outside"+overbrace(((-1)*3n))^"Inside"+overbrace(((-1)*4))^"Last"#

#color(white)((4n-1)(3n+4)) = 12n^2+16n-3n-4#

#color(white)((4n-1)(3n+4)) = 12n^2+13n-4#

FOIL can be used when multiplying any two binomials.