How do you FOIL # (x^3 – x^2)(x^3 + x^2)#?

3 Answers
Mar 3, 2018

Answer:

#(x^3-x^2)(x^3+x^2) = x^6-x^4#

Explanation:

FOIL is a mnemonic to help remember all of the combinations you need to multiply and add when finding the product of two binomials.

In our example, we find:

#(x^3-x^2)(x^3+x^2) = overbrace((x^3)(x^3))^"First"+overbrace((x^3)(x^2))^"Outside"+overbrace((-x^2)(x^3))^"Inside"+overbrace((-x^2)(x^2))^"Last"#

#color(white)((x^3-x^2)(x^3+x^2)) = x^6+color(red)(cancel(color(black)(x^5)))-color(red)(cancel(color(black)(x^5)))-x^4#

#color(white)((x^3-x^2)(x^3+x^2)) = x^6-x^4#

In fact, in this particular example, we might note that the multiplication takes the form #(A-B)(A+B)# with #A=x^3# and #B=x^2#.

In general, we have:

#(A-B)(A+B) = A^2-B^2#

This is known as the difference of squares identity.

In our particular example:

#(x^3-x^2)(x^3+x^2) = (x^3)^2-(x^2)^2 = x^6-x^4#

Mar 3, 2018

Answer:

#x^6-x^4#, or alternatively #x^2(x^3-x^2)# (factored form)

Explanation:

FOIL is extremely useful in multiplying binomials. To be able to use it, let's understand what it means first:

FOIL stands for Firsts, Outsides, Insides, Lasts, meaning we multiply the first terms, outside terms, inside terms, and last terms, respectively.

In #(x^3-x^2)(x^3+x^2)#, our terms are as follows:

  • Firsts #(x^3*x^3)= x^6#
  • Outsides #(x^3*x^2)= x^5#
  • Insides #(-x^2*x^3)= -x^5#
  • Lasts #(-x^2*x^2)= -x^4#

Thus, we have:

#x^6+x^5-x^5-x^4#

The middle terms will cancel out (as expected, because the original problem is in the difference of squares format), and we get:

#x^6-x^4#

We can factor out an #x^2# (which is essentially dividing by #x^2#). We get:

#x^2(x^3-x^2)#

Mar 3, 2018

Answer:

the answer is #x^6#-#x^4#

Explanation:

(#x^3#-#x^2#)(#x^3#+#x^2#)=

=#x^3##x^3# + #x^3##x^2# - #x^2##x^3# - #x^2##x^2#=
_ First Outer Inner __ Last
=#x^6#-#x^4#