How to use FOIL in the distributive property - PSAT Math
Card 0 of 77

Given the equation above, what is the value of
?
Given the equation above, what is the value of ?
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Use FOIL to expand the left side of the equation.


From this equation, we can solve for
,
, and
.



Plug these values into
to solve.

Use FOIL to expand the left side of the equation.
From this equation, we can solve for ,
, and
.
Plug these values into to solve.
Which of the following is equal to the expression
?
Which of the following is equal to the expression ?
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Multiply using FOIL:
First = 3x(2x) = 6x2
Outter = 3x(4) = 12x
Inner = -1(2x) = -2x
Last = -1(4) = -4
Combine and simplify:
6x2 + 12x - 2x - 4 = 6x2 +10x - 4
Multiply using FOIL:
First = 3x(2x) = 6x2
Outter = 3x(4) = 12x
Inner = -1(2x) = -2x
Last = -1(4) = -4
Combine and simplify:
6x2 + 12x - 2x - 4 = 6x2 +10x - 4
If
, what is the value of
?
If , what is the value of
?
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Remember that (a – b )(a + b ) = a 2 – b 2.
We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.
Simplify to find 9_x_2 – 16 = 2.
Adding 16 to each side gives us 9_x_2 = 18.
Remember that (a – b )(a + b ) = a 2 – b 2.
We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.
Simplify to find 9_x_2 – 16 = 2.
Adding 16 to each side gives us 9_x_2 = 18.
If
and
, then which of the following is equivalent to
?
If and
, then which of the following is equivalent to
?
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We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x2 – 2 and h(x) = x + 4. Let's find expressions for both.
g(h(x)) = g(x + 4) = 2(x + 4)2 – 2
g(h(x)) = 2(x + 4)(x + 4) – 2
In order to find (x+4)(x+4) we can use the FOIL method.
(x + 4)(x + 4) = x2 + 4x + 4x + 16
g(h(x)) = 2(x2 + 4x + 4x + 16) – 2
g(h(x)) = 2(x2 + 8x + 16) – 2
Distribute and simplify.
g(h(x)) = 2x2 + 16x + 32 – 2
g(h(x)) = 2x2 + 16x + 30
Now, we need to find h(g(x)).
h(g(x)) = h(2x2 – 2) = 2x2 – 2 + 4
h(g(x)) = 2x2 + 2
Finally, we can find g(h(x)) – h(g(x)).
g(h(x)) – h(g(x)) = 2x2 + 16x + 30 – (2x2 + 2)
= 2x2 + 16x + 30 – 2x2 – 2
= 16x + 28
The answer is 16x + 28.
We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x2 – 2 and h(x) = x + 4. Let's find expressions for both.
g(h(x)) = g(x + 4) = 2(x + 4)2 – 2
g(h(x)) = 2(x + 4)(x + 4) – 2
In order to find (x+4)(x+4) we can use the FOIL method.
(x + 4)(x + 4) = x2 + 4x + 4x + 16
g(h(x)) = 2(x2 + 4x + 4x + 16) – 2
g(h(x)) = 2(x2 + 8x + 16) – 2
Distribute and simplify.
g(h(x)) = 2x2 + 16x + 32 – 2
g(h(x)) = 2x2 + 16x + 30
Now, we need to find h(g(x)).
h(g(x)) = h(2x2 – 2) = 2x2 – 2 + 4
h(g(x)) = 2x2 + 2
Finally, we can find g(h(x)) – h(g(x)).
g(h(x)) – h(g(x)) = 2x2 + 16x + 30 – (2x2 + 2)
= 2x2 + 16x + 30 – 2x2 – 2
= 16x + 28
The answer is 16x + 28.
Simplify the expression.

Simplify the expression.
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Solve by applying FOIL:
First: 2x2 * 2y = 4x2y
Outer: 2x2 * a = 2ax2
Inner: –3x * 2y = –6xy
Last: –3x * a = –3ax
Add them together: 4x2y + 2ax2 – 6xy – 3ax
There are no common terms, so we are done.
Solve by applying FOIL:
First: 2x2 * 2y = 4x2y
Outer: 2x2 * a = 2ax2
Inner: –3x * 2y = –6xy
Last: –3x * a = –3ax
Add them together: 4x2y + 2ax2 – 6xy – 3ax
There are no common terms, so we are done.
The sum of two numbers is
. The product of the same two numbers is
. If the two numbers are each increased by one, the new product is
. Find
in terms of __
_.
The sum of two numbers is . The product of the same two numbers is
. If the two numbers are each increased by one, the new product is
. Find
in terms of __
_.
Tap to see back →
Let the two numbers be x and y.
x + y = s
xy = p
(x + 1)(y + 1) = q
Expand the last equation:
xy + x + y + 1 = q
Note that both of the first two equations can be substituted into this new equation:
p + s + 1 = q
Solve this equation for q – p by subtracting p from both sides:
s + 1 = q – p
Let the two numbers be x and y.
x + y = s
xy = p
(x + 1)(y + 1) = q
Expand the last equation:
xy + x + y + 1 = q
Note that both of the first two equations can be substituted into this new equation:
p + s + 1 = q
Solve this equation for q – p by subtracting p from both sides:
s + 1 = q – p
Expand the expression:
$(x^{3}$-4x)(6 + $12x^{2}$)
Expand the expression:
$(x^{3}$-4x)(6 + $12x^{2}$)
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When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.
$(x^{3}$-4x)(6 + $12x^{2}$)
$6x^{3}$$+12x^{5}$$-24x-48x^{3}$
$-42x^{3}$$+12x^{5}$-24x
$12x^{5}$$-42x^{3}$-24x
When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.
$(x^{3}$-4x)(6 + $12x^{2}$)
$6x^{3}$$+12x^{5}$$-24x-48x^{3}$
$-42x^{3}$$+12x^{5}$-24x
$12x^{5}$$-42x^{3}$-24x
Expand the following expression:
$(4x+2)(x^2$-2)
Expand the following expression:
$(4x+2)(x^2$-2)
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$(4x+2)(x^2$-2)=(4xtimes $x^2$)+(4xtimes -2)+(2times $x^2$) +(2times -2)
Which becomes
$4x^3$$-8x+2x^2$-4
Or, written better
$4x^3$$+2x^2$-8x-4
$(4x+2)(x^2$-2)=(4xtimes $x^2$)+(4xtimes -2)+(2times $x^2$) +(2times -2)
Which becomes
$4x^3$$-8x+2x^2$-4
Or, written better
$4x^3$$+2x^2$-8x-4
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Expand and simplify the expression.

Expand and simplify the expression.
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We can solve by FOIL, then distribute the
. Since all terms are being multiplied, you will get the same answer if you distribute the
before using FOIL.
First: 
Inside: 
Outside: 
Last: 
Sum all of the terms and simplify. Do not forget the
in front of the quadratic!


Finally, distribute the
.

We can solve by FOIL, then distribute the . Since all terms are being multiplied, you will get the same answer if you distribute the
before using FOIL.
First:
Inside:
Outside:
Last:
Sum all of the terms and simplify. Do not forget the in front of the quadratic!
Finally, distribute the .
If
,
, and
, then 
If ,
, and
, then
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To find what
equals, you must know how to multiply
times
, or, you must know how to multiply binomials. The best way to multiply monomials is the FOIL (first, outside, inside, last) method, as shown below:

Multiply the First terms

Multiply the Outside terms:

Multiply the Inside terms:

Note: this step yields a negative number because the product of the two terms is negative.
Multiply the Last terms:

Note: this step yields a negative number too!
Putting the results together, you get:

Finally, combine like terms, and you get:

To find what equals, you must know how to multiply
times
, or, you must know how to multiply binomials. The best way to multiply monomials is the FOIL (first, outside, inside, last) method, as shown below:
Multiply the First terms
Multiply the Outside terms:
Multiply the Inside terms:
Note: this step yields a negative number because the product of the two terms is negative.
Multiply the Last terms:
Note: this step yields a negative number too!
Putting the results together, you get:
Finally, combine like terms, and you get:

Given the equation above, what is the value of
?
Given the equation above, what is the value of ?
Tap to see back →

Use FOIL to expand the left side of the equation.


From this equation, we can solve for
,
, and
.



Plug these values into
to solve.

Use FOIL to expand the left side of the equation.
From this equation, we can solve for ,
, and
.
Plug these values into to solve.
Which of the following is equal to the expression
?
Which of the following is equal to the expression ?
Tap to see back →
Multiply using FOIL:
First = 3x(2x) = 6x2
Outter = 3x(4) = 12x
Inner = -1(2x) = -2x
Last = -1(4) = -4
Combine and simplify:
6x2 + 12x - 2x - 4 = 6x2 +10x - 4
Multiply using FOIL:
First = 3x(2x) = 6x2
Outter = 3x(4) = 12x
Inner = -1(2x) = -2x
Last = -1(4) = -4
Combine and simplify:
6x2 + 12x - 2x - 4 = 6x2 +10x - 4
If
, what is the value of
?
If , what is the value of
?
Tap to see back →
Remember that (a – b )(a + b ) = a 2 – b 2.
We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.
Simplify to find 9_x_2 – 16 = 2.
Adding 16 to each side gives us 9_x_2 = 18.
Remember that (a – b )(a + b ) = a 2 – b 2.
We can therefore rewrite (3_x –_ 4)(3_x_ + 4) = 2 as (3_x_ )2 – (4)2 = 2.
Simplify to find 9_x_2 – 16 = 2.
Adding 16 to each side gives us 9_x_2 = 18.
If
and
, then which of the following is equivalent to
?
If and
, then which of the following is equivalent to
?
Tap to see back →
We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x2 – 2 and h(x) = x + 4. Let's find expressions for both.
g(h(x)) = g(x + 4) = 2(x + 4)2 – 2
g(h(x)) = 2(x + 4)(x + 4) – 2
In order to find (x+4)(x+4) we can use the FOIL method.
(x + 4)(x + 4) = x2 + 4x + 4x + 16
g(h(x)) = 2(x2 + 4x + 4x + 16) – 2
g(h(x)) = 2(x2 + 8x + 16) – 2
Distribute and simplify.
g(h(x)) = 2x2 + 16x + 32 – 2
g(h(x)) = 2x2 + 16x + 30
Now, we need to find h(g(x)).
h(g(x)) = h(2x2 – 2) = 2x2 – 2 + 4
h(g(x)) = 2x2 + 2
Finally, we can find g(h(x)) – h(g(x)).
g(h(x)) – h(g(x)) = 2x2 + 16x + 30 – (2x2 + 2)
= 2x2 + 16x + 30 – 2x2 – 2
= 16x + 28
The answer is 16x + 28.
We are asked to find the difference between g(h(x)) and h(g(x)), where g(x) = 2x2 – 2 and h(x) = x + 4. Let's find expressions for both.
g(h(x)) = g(x + 4) = 2(x + 4)2 – 2
g(h(x)) = 2(x + 4)(x + 4) – 2
In order to find (x+4)(x+4) we can use the FOIL method.
(x + 4)(x + 4) = x2 + 4x + 4x + 16
g(h(x)) = 2(x2 + 4x + 4x + 16) – 2
g(h(x)) = 2(x2 + 8x + 16) – 2
Distribute and simplify.
g(h(x)) = 2x2 + 16x + 32 – 2
g(h(x)) = 2x2 + 16x + 30
Now, we need to find h(g(x)).
h(g(x)) = h(2x2 – 2) = 2x2 – 2 + 4
h(g(x)) = 2x2 + 2
Finally, we can find g(h(x)) – h(g(x)).
g(h(x)) – h(g(x)) = 2x2 + 16x + 30 – (2x2 + 2)
= 2x2 + 16x + 30 – 2x2 – 2
= 16x + 28
The answer is 16x + 28.
Simplify the expression.

Simplify the expression.
Tap to see back →
Solve by applying FOIL:
First: 2x2 * 2y = 4x2y
Outer: 2x2 * a = 2ax2
Inner: –3x * 2y = –6xy
Last: –3x * a = –3ax
Add them together: 4x2y + 2ax2 – 6xy – 3ax
There are no common terms, so we are done.
Solve by applying FOIL:
First: 2x2 * 2y = 4x2y
Outer: 2x2 * a = 2ax2
Inner: –3x * 2y = –6xy
Last: –3x * a = –3ax
Add them together: 4x2y + 2ax2 – 6xy – 3ax
There are no common terms, so we are done.
The sum of two numbers is
. The product of the same two numbers is
. If the two numbers are each increased by one, the new product is
. Find
in terms of __
_.
The sum of two numbers is . The product of the same two numbers is
. If the two numbers are each increased by one, the new product is
. Find
in terms of __
_.
Tap to see back →
Let the two numbers be x and y.
x + y = s
xy = p
(x + 1)(y + 1) = q
Expand the last equation:
xy + x + y + 1 = q
Note that both of the first two equations can be substituted into this new equation:
p + s + 1 = q
Solve this equation for q – p by subtracting p from both sides:
s + 1 = q – p
Let the two numbers be x and y.
x + y = s
xy = p
(x + 1)(y + 1) = q
Expand the last equation:
xy + x + y + 1 = q
Note that both of the first two equations can be substituted into this new equation:
p + s + 1 = q
Solve this equation for q – p by subtracting p from both sides:
s + 1 = q – p
Expand the expression:
$(x^{3}$-4x)(6 + $12x^{2}$)
Expand the expression:
$(x^{3}$-4x)(6 + $12x^{2}$)
Tap to see back →
When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.
$(x^{3}$-4x)(6 + $12x^{2}$)
$6x^{3}$$+12x^{5}$$-24x-48x^{3}$
$-42x^{3}$$+12x^{5}$-24x
$12x^{5}$$-42x^{3}$-24x
When using FOIL, multiply the first, outside, inside, then last expressions; then combine like terms.
$(x^{3}$-4x)(6 + $12x^{2}$)
$6x^{3}$$+12x^{5}$$-24x-48x^{3}$
$-42x^{3}$$+12x^{5}$-24x
$12x^{5}$$-42x^{3}$-24x
Expand the following expression:
$(4x+2)(x^2$-2)
Expand the following expression:
$(4x+2)(x^2$-2)
Tap to see back →
$(4x+2)(x^2$-2)=(4xtimes $x^2$)+(4xtimes -2)+(2times $x^2$) +(2times -2)
Which becomes
$4x^3$$-8x+2x^2$-4
Or, written better
$4x^3$$+2x^2$-8x-4
$(4x+2)(x^2$-2)=(4xtimes $x^2$)+(4xtimes -2)+(2times $x^2$) +(2times -2)
Which becomes
$4x^3$$-8x+2x^2$-4
Or, written better
$4x^3$$+2x^2$-8x-4
Tap to see back →