Article: http://www.academyofone.org/014-adding-subtracting-polynomials-algebra-1-13/ Visit: http://academyofone.org AoO Prime: http://www.academyofone.org/register-2/ Royalty free music licensed by http://www.stockmusic.net Royalty free photos licensed by http://www.bigstockphoto.com Script: Alright, so now that we know and understand polynomials, we can apply our mathematical operators to them. What’s going on everybody! My name is Jack Jenkins and this is Academy of One. Today we are adding and subtracting polynomials. So to add or subtract polynomials, all you need to do is combine the like terms. That’s it! It’s really simple and we’ll do some examples. So what’s seven x plus two in parenthesis plus eighteen x plus four in parenthesis. I’ll give you a minute to do this. Welcome back from the problem. To do this all we need to do is combine the like terms. Seven x plus eighteen x is twenty-five x. The two and the four will combine to form six so the answer to seven x plus two in Parenthesis plus eighteen x plus four in parenthesis is twenty-five x plus six. Now here’s a bonus question, we are about to evaluate the algebraic expression. The variable x is three, now do the problem quickly! Three… Two… One… Alright so twenty-five times three is seventy-five plus six is eighty-one. Great job! Now let’s do yet another problem. Seven y to the power of four plus negative eight y to the power of three in parenthesis plus two to the power of two plus zero to the power of one plus nine in parenthesis. Solve! Now this was actually a trick question. There wasn’t any polynomials to combine. I just wanted to see if you guys can simplify this expression. The seven y to the power of four is ok. Plus negative eight y to the power of three is just minus eight y to the power of three. Two to the power of two is fine, but the zero the power of one… Nope! Exponents is the same as multiplying a number to itself and anything times zero is just zero so we don’t need to write that…. Also x to the power of one is just gonna be x. The plus nine is great though. So the answer to Seven y to the power of four plus negative eight y to the power of three in parenthesis plus two to the power of two plus zero to the power of one plus nine in parenthesis is seven y to the power of four minus negative eight y to the power of three plus two to the power of two plus nine. Last question, this is gonna be a subtraction problem. Six q to the power of two plus eight v to the power of two in parenthesis minus nine q to the power of two minus two v to the power of two in parenthesis. Solve it. What I like to do is rewrite the expression and change the minus sign into a positive sign. This means we have to flip the operator in-front of the numbers to the right of the minus sign. The problem shall be rewritten as six q to the power of two plus eight v to the power of two in parenthesis plus negative nine q to the power of two plus two v to the power of two in parenthesis. After we rewrite the problem you just have to combine the like terms. Six q to the power of two plus negative nine q to the power of two is negative three q to the power of two. The eight v minus two v will combine into six v. The answer to the problem is negative three q to the power of two plus six v to the power of two. Great job. Well that was interesting! Adding and subtracting polynomials is easy… But you want to know what’s a little harder? Multiplying polynomials.

• published: 29 Oct 2014
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