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

Question

Savannah Anderson · Accepted Answer

#(x+3) * (x+4) = x^2 + 7*x + 12#

The rigorous but lengthy explanation includes the commutative law for addition and multiplication as well as the distributive law for multiplying a sum of two numbers by a third.

The distributive law of multiplication states: #a*(b+c)=a*b+a*c#

The commutative law of addition states: #a+b = b+a#

The commutative law of multiplication states" #a*b = b*a#

All of the aforementioned laws are applied when solving this problem, usually without being specifically mentioned.

Using the distributive law described above for #a=x+3#, #b=x# and #c=4#, we obtain: #(x+3) * (x+4) = (x+3) * x + (x+3) * 4#

Now let's use the commutative law of multiplication to rewrite it as #(x+3) * (x+4) = x * (x+3) + 4 * (x+3)#

Now use the distributive law twice, first for #a=x#, #b=x# and #c=3# obtaining #x * (x+3) = x*x + x*3# and another time for #a=4#, #b=x# and #c=3# obtaining #4 * (x+3) = 4*x + 4*3#

Let's use these equalities in the original expression: #(x+3) * (x+4) = x*x + x*3 + 4*x + 4*3#

Further simplification involves the following transformations: #x*x = x^2# (just for convenience) Using commutative and distributive laws, #x*3+4*x = x*3+x*4=x*(3+4)=x*7=7*x# Obviously, #4*3=12#

Our expression looks now as follows: #(x+3) * (x+4) = x^2 + 7*x + 12# This is the final representation of the original expression as a polynomial of the second degree.

Naturally, the majority of students perform all these transformations without even considering the laws upon which they are based, nor do they write an intermediary result. However, these abilities are developed through repeated practice—practice makes perfect.

Jace Anderson · Answer

undefined To multiply $ (x+3)(x+4) $, you can use the distributive property or the FOIL method:

Using the distributive property:
$ (x+3)(x+4) = x(x+4) + 3(x+4) $

Expanding each term:
$ = x^2 + 4x + 3x + 12 $

Combining like terms:
$ = x^2 + 7x + 12 $

Using the FOIL method:
FOIL stands for First, Outer, Inner, Last.
$ (x+3)(x+4) = x \cdot x + x \cdot 4 + 3 \cdot x + 3 \cdot 4 $

Expanding each term:
$ = x^2 + 4x + 3x + 12 $

Combining like terms:
$ = x^2 + 7x + 12 $

So, $ (x+3)(x+4) = x^2 + 7x + 12 $.