Unit 2:Dividing Polynomials!

This is an example of a problem, following the rules of dividing polynomials.

Step 1:(x+4)(x+3)(x+2)(x+1) / x+2

My first step is to come up with my original and writing my original equation. My goal is to solve this equation using the rules of Dividing Polynomials. I will have to solve it and divide it, until there is a final equation. At the end when I complete the division there should be no remainder.

Step 2: x^2+3x+4x+12

My next step is to FOIL the first set of x and values and the second set of x and values. You can FOIL the first set of equations: (x+4)(x+3) by first multiplying the two x’s together and getting x^2. After that you multiply the first x by the 3 in order to get the 3x and then you multiply the 4 by the second x in order to get 4x. Last you have to multiply the 4 and 3 by each other in order to get the last part of the equation which is the twelve. Once this is done you end up with the equation x^2+3x+4x+12.

Step 3: x^2+1x+2x+2

After that my next step is to complete the same steps using the last two equations. I start off by multiplying the x terms in order to get x^2. I then multiply the 1 and the x in order to get 1x. After doing that you have to multiply the 2 and the x in order to get 2x. Last you have to multiply the 1 and the 2 together so that you can get the final number in the equation which is 2. This then gives you the final equation of x^2+1x+2x+2.

Step 4: x^2+7x+12 & x^2+3x+2

My next step is to add the like terms in the both equations, In the first one I had to add the 4 and the 3 due to the fact that they were both coefficients in front of the x values. After adding those two terms I ended up with 7x, so I then had to plug that back into the equation leaving me with my new equation: x^2+7x+12. After doing that with the first equation, I had to then repeat the same with the second equation. I had to find the two like terms which were 1x and 2x and then I had to add their coefficients. After adding their coefficients I ended up with 3x and then I inserted that into the equation making my new equation:x^2+3x+2.

Step 5: (x^2+7x+12)( x^2+3x+2)

After forming both of my new equations, I then had to multiply them together using the box method. I completed this by multiplying each term in the first equation by the second equation.

The first step was to multiply the two x^2’s by each other which gave me x^4. The exponent became a four due to the rules of multiplying numbers with exponents attached, which is that if you multiply then you must add the exponents attached to the numbers in front of the x terms. My next step was to multiply the 7x by x^2 which gave me 7x^3. After completing that I multiplied 12 by the x^2 and got 12x^2. Next, I did the same with the second row of terms, but instead I multiplied each term by 3x. My first step was to multiply the 3x by the x2in order to get 3x^3. Next, I multiplied the 7x by the 3x in order to get 21x^2. The last step for that row was to multiply 12 by 3x and when I did that I got the answer of 36x. For the last row my first step was to multiply the 2 by the x2 and once I did that I came up with 2x^2. The next step was for me to multiply the 2 by the 7x which gave me 14x. My final step for the multiplication was to multiply the two final terms: 2 and 12 together in order to get the answer of 24. This completes the multiplication part of the solving for the equation.

Step 6: x^4+10x^3+35x^2+50x+24

My next step was to write my new polynomial which I found by adding up all of my like terms from the multiplication process. I added each term that had the same exponent value together and once I did I got the polynomial shown above.

Step 7: x+2 / x^4+10x^3+35x^2+50x+24

The final step of my equation was to follow the rules of dividing polynomials and divide the new equation by x+2.

When I first start my division I have to see what times the x could give me the values in the equation. The first value is an x^4, so I had to figure out what times the x could give me that. I came to the conclusion that x3would be the answer due to the exponent rules so I put that on top of the radical. Next I moved to multiplying the 2 by the x^3. When I multiplied these two terms, I got 2x^3. In order to move on to the next step I had to subtract the 2x^3 from the 10x^3 that was in the original equation. When completing the subtraction I ended up with 8x^3. Once that part was complete I had to figure out what times the x could give me that, so I determined that the number was 8x^2. After getting that I added it to the top of the radical. Next I had to carry down the 35x^2. I then had to multiply the 8x^2 by the 2 in order to find what I needed to subtract/ add from the 35x^2.After multiplying the two together, I was left with 16x^2 which I then subtracted from the 35x^2 in order to get 19x^2. After that, I had to carry down the 50x and multiply the 19x^2 and the 2 in order to get 38x which was the term that I had to subtract from the 50x. Once I did that, I had to figure out what times x could give me the 19x^2, so when I did I determined that the answer was 19x and I added it above my radical. After subtracting that from the 50x I was left with 12x and from there I had to figure out what times the x could give me 12x so when I saw that it was 12, I moved that above my radical and then I carried down the 24. Lastly I multiplied the 12 by the 2 and got 24. When I subtracted the two 24’s I got a remainder of 0, which meant that x+2 is a factor of the original equation.