Distributive Property - PSAT Math

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

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.

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.

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.

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

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

$(4x+2)(x^2$-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

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 .

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:

Use FOIL to expand the left side of the equation.

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

Plug these values into to solve.

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

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.

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.

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.

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

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

$(4x+2)(x^2$-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