What is the standard form of #y= (x+3) (4x+1) #?

1 Answer
George C.
Dec 27, 2015

#y = 4x^2+13x+3#

Explanation:

Use FOIL to multiply out...

#(x+3)(4x+1) = stackrel "First" overbrace(x*4x) + stackrel "Outside"overbrace(x*1) + stackrel "Inside" overbrace(3*4x) + stackrel "Last" overbrace(3*1)#

#=4x^2+x+12x+3 = 4x^2+13x+3#

Standard form has individual terms in descending order of degree of #x#. If the binomial factors are expressed in the form #(ax+b)# then the result of FOIL will be in the right order, just requiring the combination of the middle terms.

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