\(\displaystyle (\sqrt5+\sqrt3)^2=(\sqrt5)^2+2\sqrt5\sqrt3+(\sqrt3)^2=5+2\sqrt{15}+3=8+2\sqrt{15}\)

Remember that (a – b )(a + b ) = a ² – b ².

We can therefore rewrite (3x – 4)(3x + 4) = 2 as (3x )² – (4)² = 2.

Simplify to find 9x² – 16 = 2.

Adding 16 to each side gives us 9x² = 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.

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.

\(\displaystyle \dpi{100} \small (x^{3}-4x)(6 + 12x^{2})\)

\(\displaystyle \dpi{100} \small 6x^{3}+12x^{5}-24x-48x^{3}\)

\(\displaystyle \dpi{100} \small -42x^{3}+12x^{5}-24x\)

\(\displaystyle \dpi{100} \small 12x^{5}-42x^{3}-24x\)

\(\displaystyle (4x+2)(x^2-2)=(4x\times x^2)+(4x\times -2)+(2\times x^2) +(2\times -2)\)

Which becomes

\(\displaystyle 4x^3-8x+2x^2-4\)

Or, written better

\(\displaystyle 4x^3+2x^2-8x-4\)

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

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.

\(\displaystyle (6x - 2) (3x + 5) = ax^2 + dx + k\)

Use FOIL to expand the left side of the equation.

\(\displaystyle (6x-2)(3x+5)=18x^2+30x-6x-10=18x^2+24x-10\)

\(\displaystyle 18x^2+24x-10=ax^2+dx+k\)

From this equation, we can solve for \(\displaystyle a\), \(\displaystyle d\), and \(\displaystyle k\).

\(\displaystyle 18x^2=ax^2\rightarrow a=18\)

\(\displaystyle 24x=dx\rightarrow d=24\)

\(\displaystyle -10=k\)

Plug these values into \(\displaystyle a-d+k\) to solve.

\(\displaystyle a-d+k=(18)-(24)+(-10)=-16\)

\(\displaystyle 4(x-5)(2x+10)\)

We can solve by FOIL, then distribute the \(\displaystyle \small 4\). Since all terms are being multiplied, you will get the same answer if you distribute the \(\displaystyle \small 4\) before using FOIL.

First: \(\displaystyle x*2x=2x^2\)

Inside: \(\displaystyle -5*2x=-10x\)

Outside: \(\displaystyle x*10=10x\)

Last: \(\displaystyle -5*10=-50\)

Sum all of the terms and simplify. Do not forget the \(\displaystyle \small 4\) in front of the quadratic!

\(\displaystyle 4(2x^2-10x+10x-50)\)

\(\displaystyle 4(2x^2-50)\)

Finally, distribute the \(\displaystyle \small 4\).

\(\displaystyle 8x^2-200\)