How do you multiply #(3x + 4)(2x + 3)#?

1 Answer
Jul 11, 2016

Answer:

#6x^2+17x+12#

Explanation:

Most simple transformations of algebraic expressions are based on a set of laws:
Commutative law of addition:
#a+b = b+a#
Associative law of addition:
#(a+b)+c = a+(b+c)#
Commutative law of multiplication:
#a*b = b*a#
Associative law of multiplication:
#(a*b) * c = a * (b*c)#
Distributive law of multiplication over addition:
#(a+b)*c = a*c+b*c#
and, derived from it and from commutative law, the second form:
#c*(a+b)=c*a+c*b#
(The distributive law is often referred to as "opening the parenthesis")

Also, let's recall the rule of precedence: multiplication and division should be performed first, left to right, then addition and subtraction are performed with intermediary results, also left to right.

Using the distributive law in its first form for
#a = 3x#,
#b = 4#
#c = 2x+3#
we can write:
#(3x+4)(2x+3) = 3x(2x+3) + 4(2x+3)#

The first component, using the same distributive law in its second form, is
#3x(2x+3) = 3x*2x+3x*3#
which, using the commutative and associative laws of multiplication, equals to
# = 3*2*x*x+3*3*x = 6x^2+9x#

The second component, using the same laws can be transformed into
#4(2x+3) = 4*2x+4*3 = 8x+12#

Adding together these two components in their final forms, we get:
#6x^2+9x+8x+12#

This can be simplified further using the distributive law to two elements with #x#:
#9x+8x = (9+8)x = 17x#

That results in the final form for our expression:
#6x^2+17x+12#