Step 1: Find factors of 44:

\(\displaystyle (1,44), (2,22), (4,11)\)

Step 2: Find which pair of factors can give me the middle number. We will choose \(\displaystyle (4,11)\).

Step 3: Using \(\displaystyle 4\) and \(\displaystyle 11\), we need to get \(\displaystyle -7\). The only way to get \(\displaystyle -7\) is if I have \(\displaystyle -11\) and \(\displaystyle 4\).

Step 4: Write the factored form of that trinomial:

\(\displaystyle (x-11)(x+4)\)

Step 5: To solve for x, you set each parentheses to \(\displaystyle 0\):

\(\displaystyle (x-11)=0\rightarrow x=11\)
\(\displaystyle (x+4)=0\rightarrow x=-4\)

The solutions to this equation are \(\displaystyle x=11\) and \(\displaystyle x=-4\).

Step 1: Factor by pairs:

\(\displaystyle x^3-4x^2+x-4=0\)

\(\displaystyle x^2(x-4)+1(x-4)=0\)

Step 2: Re-write the factorization:  

\(\displaystyle (x^2+1)(x-4)=0\)

Step 3: Solve for x:

\(\displaystyle (x^2+1)=0;(x-4)=0\)
\(\displaystyle \\ x^2=-1;x=4 \\\sqrt{x^2}=\pm\sqrt{-1}; x=4 \\x=\pm i;x=4\)

Step 1: Find two numbers that multiply to \(\displaystyle -6\) and add to \(\displaystyle 5\).

We will choose \(\displaystyle 6,-1\).

Step 2: Factor using the numbers we chose:

\(\displaystyle (x+6)(x-1)=0\)

Step 3: Solve each parentheses for each value of x..

\(\displaystyle (x+6)=0 \implies x=-6\)
\(\displaystyle (x-1)=0\implies x=1\)

Based upon the fundamental theorem of algebra, we know that there must exist 3 roots for this polynomial based upon its' degree of 3. 

To solve for the roots, we use factor by grouping: 

\(\displaystyle x^3+5x^2-3x-15=0\)

First group the terms into two binomials: 

\(\displaystyle (x^3+5x^2)(-3x-15)=0\)

Then take out the greatest common factor from each group: 

\(\displaystyle x^2(x+5)-3(x+5)=0\)

Now we see that the leftover binomial is the greatest common factor itself: 

\(\displaystyle (x+5)\ is\ common\ to\ both\ terms\ so\ we\ can\ take\ it\ out:\)

\(\displaystyle (x+5)(x^2-3)=0\)

We set each binomial equal to zero and solve: 

\(\displaystyle x+5=0\ so\ x=-5\)

\(\displaystyle x^2-3=0;\ so\ x^2=3\ which\ means:\)

\(\displaystyle x=\pm\sqrt{3}\ after\ taking\ the\ square\ root\ of\ both\ sides.\)

\(\displaystyle The\ 3\ roots\ are\ -5, {\sqrt{3}},\ and\ -\sqrt{3}\)

In order to find the roots for the polynomial we must first put it in Standard Form by decreasing exponent: 

\(\displaystyle f(x)=x^3+21-7x^2-3x\)

\(\displaystyle x^3-7x^2-3x+21\)

Now we can use factor by grouping, we start by grouping the 4 terms into 2 binomials: 

\(\displaystyle (x^3-7x^2)(-3x+21)\)

We now take the greatest common factor out of each binomial: 

\(\displaystyle x^2(x-7)-3(x-7)\)

We can see that each term now has the same binomial as a common factor, so we simplify to get: 

\(\displaystyle (x-7)(x^2-3)\)

To find all of the roots, we set each factor equal to zero and solve: 

\(\displaystyle x-7=0\ which\ gives\ us\ x=7\)

\(\displaystyle x^2-3=0\ which\ gives\ us\ x^2=3\ which\ then\ yields\ x=\pm \sqrt{3}\)

\(\displaystyle The\ roots\ are\ 7,\ \sqrt{3}\ and\ -\sqrt{3}\)

\(\displaystyle If\ the\ complex\ root\ a+bi\ exists, then\ a-bi\ must\ also\ exist\ as\ a\ root.\)

\(\displaystyle If\ the\ irrational\ root\ a+\sqrt{b}\ exists,\ then\ the\ root\ a-\sqrt{b}\ must\ also\ exist.\)

\(\displaystyle Given\ that\ the\ above\ statements\ always\ hold\ true\ we\ know\ that:\)

\(\displaystyle If\ -\sqrt{5}\ exists\ as\ a\ root\, then\ +\sqrt{5}\ also\ exists\ as\ a\ root.\)

\(\displaystyle Also,\ if\ 3+2i\ exists\ as\ a\ root,\ then\ 3-2i\ must\ also\ exist\ as\ a\ root.\)

 \(\displaystyle This\ gives\ us\ an\ additional\ 2\ roots\ above\ the\ 3\ roots\ we\ already\ know.\)

\(\displaystyle There\ will\ be\ a\ total\ of\ 5\ roots.\)

Step 1: Find two numbers that multiply to \(\displaystyle \large -10\) and add to \(\displaystyle \large -3\)...

We will choose \(\displaystyle \large -5, 2\)

Check: 

\(\displaystyle \large (5)(-2)=-10\)
\(\displaystyle \large -5+2=-3\)

We have the correct numbers...

Step 2: Factor the polynomial...

\(\displaystyle \large x^2-3x-10=(x-5)(x+2)\)

Step 3: Set the parentheses equal to zero to get the roots...

\(\displaystyle \large (x-5)=0\implies x=5\)

\(\displaystyle \large (x+2)=0 \implies x=-2\)

So, the roots are \(\displaystyle \large \large 5, -2\).

In order to find the roots of the polynomial we must factor by grouping: 

\(\displaystyle x^3+5x^2-4x-20\)

Group into two binomials:

\(\displaystyle (x^3+5x^2)(-4x-20)\)

Take out the greatest common factor from each binomial: 

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

We can now see that each term has a common binomial factor: 

\(\displaystyle (x+5)(x^2-4)\) 

We set each factor equal to zero and solve to obtain the roots: 

\(\displaystyle x+5=0\ and\ x^2-4=0\)

\(\displaystyle x=-5\ and\ x^2=4\)

\(\displaystyle x=-5\ and\ \sqrt{x^2}=\sqrt{4}\ so\ x=\pm2\)

\(\displaystyle The\ roots\ are\ -5,\ 2\ and\ -2\)

Step 1: Evaluate \(\displaystyle (x+2)^2\).

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

Step 2. Evaluate \(\displaystyle (x+2)^3\)

From the previous step, we already know what \(\displaystyle (x+2)^2\) is. 

\(\displaystyle (x+2)^3\) is just multiplying by another \(\displaystyle (x+2)\)

\(\displaystyle (x^2+4x+4)(x+2)=x^3+2x^2+4x^2+8x+4x+8\)
\(\displaystyle =x^3+6x^2+12x+8\)

Step 3: Evaluate \(\displaystyle (x+2)^4\). 

\(\displaystyle (x+2)^4=(x+2)^3(x+2)\)

\(\displaystyle (x^3+6x^2+12x+8)(x+2)\)
\(\displaystyle =x^4+2x^3+6x^3+12x^2+12x^2+24x+8x+16\)
\(\displaystyle =x^4+8x^3+24x^2+32x+16\)

The expansion of \(\displaystyle (x+2)^4\) is \(\displaystyle x^4+8x^3+24x^2+32x+16\)

Solution:

We can look at Pascal's Triangle, which is a quick way to do Binomial Expansion. We read each row (across, left to right)

For the first row, we only have a constant.
For the second row, we get \(\displaystyle x+constant\).
...
For the 7th row, we will start with an \(\displaystyle x^6\) term and end with a constant.

Step 1: We need to locate the 7th row of the triangle and write the numbers in that row out.

The 7th row is \(\displaystyle 1,6,15,20,15,6,1\).

Step 2: If we translate the 7th row into an equation, we get:

 \(\displaystyle x^6+6x^5+15x^4+20x^3+15x^2+6x+1\). This is the solution.