algebra help
- Thread starter DoZeRxXx
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Eh...that's what the carat button is for (^). It's not just for ^_^;
so....
X^2 is a monomial. It has 1 variable raised to a single power. 52*X^4 is another example.
X^2 + X^3 is a binomial. It has the varible raised to 2 different powers. Thus bi. Another example is X + 1.
X^2 + 2X + 1 is a polynomial. It has the variable (X) raised to 3 different powers (2, 1, and 0). It can have the variable raised to a bunch of different powers. So, you get X^4 + 32X^3 + 52X^2 + 18X + 12.5 as a different polynomial. Anything with a variable raised to more than 2 different exponents is a polynomial. Thus poly.
Now, a little lesson on "doing."
Addition.
You add the coefficients of variables that are raised to the same exponent.
X^2 + 3X^2 = 4X^2
(plz don't add X's and Y's. X + Y does not equal 2X or 2Y)
Multiplication.
You FOIL. This is used for binomials, the most common sort of polynomial multiplication. You multiply the first terms in the two chunks, then outside, inside, and last. You add them together. First, Outside, Inside, Last. Foil. If you need to multiply polynomials with more than 2 terms, you do basically the same thing. Just remember that every term in the 1st must be multiplied by every term in the 2nd. Then add.
(X+1)(X+1)
X*X + X*1 + X*1 + 1*1 = X^2 + 2X + 1
Division.
No way in hell I'm gonna be able to explain this by typing. You'll have to look it up in the book. I'll put a couple easy, semi-followable examples here.
Bleh. Paint sucks.
Hope I showed you something you needed. Good luck on the quiz 👍
[edit] bah. Now he went away. What's the use....
so....
X^2 is a monomial. It has 1 variable raised to a single power. 52*X^4 is another example.
X^2 + X^3 is a binomial. It has the varible raised to 2 different powers. Thus bi. Another example is X + 1.
X^2 + 2X + 1 is a polynomial. It has the variable (X) raised to 3 different powers (2, 1, and 0). It can have the variable raised to a bunch of different powers. So, you get X^4 + 32X^3 + 52X^2 + 18X + 12.5 as a different polynomial. Anything with a variable raised to more than 2 different exponents is a polynomial. Thus poly.
Now, a little lesson on "doing."
Addition.
You add the coefficients of variables that are raised to the same exponent.
X^2 + 3X^2 = 4X^2
(plz don't add X's and Y's. X + Y does not equal 2X or 2Y)
Multiplication.
You FOIL. This is used for binomials, the most common sort of polynomial multiplication. You multiply the first terms in the two chunks, then outside, inside, and last. You add them together. First, Outside, Inside, Last. Foil. If you need to multiply polynomials with more than 2 terms, you do basically the same thing. Just remember that every term in the 1st must be multiplied by every term in the 2nd. Then add.
(X+1)(X+1)
X*X + X*1 + X*1 + 1*1 = X^2 + 2X + 1
Division.
No way in hell I'm gonna be able to explain this by typing. You'll have to look it up in the book. I'll put a couple easy, semi-followable examples here.
Bleh. Paint sucks.
Hope I showed you something you needed. Good luck on the quiz 👍
[edit] bah. Now he went away. What's the use....
- 658
There's also another way you can do it called synthetic division. I find it easier. It requires some simplification for some problems but after that it's easy as crap. I'm not gonna explain. I suck at typing. Just ask your teacher or look in the book. That's what I do. Anyhoo, I think it's much much easier than the long division Purple Platypus was doing.
- 1,645
Synthetic division is where you find an x-intercept of an equation (Where Y=0) , and use it to factor out that number in the non-expanded version.
Example:
Say you have the equation X^2 + 4X + 3 which can be factored into (X+3)(X+1). The do something like this:
You always drop the first number down and then it's just muliplying the factor (-3) by the number that is at the bottom. Then you take that number and put it in the next slot. You can see how this works pretty easily. The numbers you get at the bottom are the coefficients of the equation one factor lower than the original.
so your new equation would be 1X + 1.
If your synthetic division doesn't equal 0 at the end, then you've messed up.
Hope that helps
Darn, the thing i drew didnt work, ill do it quick.
Example:
Say you have the equation X^2 + 4X + 3 which can be factored into (X+3)(X+1). The do something like this:
You always drop the first number down and then it's just muliplying the factor (-3) by the number that is at the bottom. Then you take that number and put it in the next slot. You can see how this works pretty easily. The numbers you get at the bottom are the coefficients of the equation one factor lower than the original.
so your new equation would be 1X + 1.
If your synthetic division doesn't equal 0 at the end, then you've messed up.
Hope that helps
Darn, the thing i drew didnt work, ill do it quick.
- 658
the_cobbinatorSynthetic division is where you find an x-intercept of an equation (Where Y=0) , and use it to factor out that number in the non-expanded version.
Example:
Say you have the equation X^2 + 4X + 3 which can be factored into (X+3)(X+1). The do something like this:
You always drop the first number down and then it's just muliplying the factor (-3) by the number that is at the bottom. Then you take that number and put it in the next slot. You can see how this works pretty easily. The numbers you get at the bottom are the coefficients of the equation one factor lower than the original.
so your new equation would be 1X + 1.
If your synthetic division doesn't equal 0 at the end, then you've messed up.
Hope that helps
Darn, the thing i drew didnt work, ill do it quick.
Nope, if your division doesn't equal division and you know that you've done everything right, then you have a remainder. At that point you put the remainder over.....damn...it goes over something. I got the book out and now i'm confused. Sorry. No wonder I made a 77 on the last quiz. Anyhoo, I do know this, that in the equation that you just gave can be solved in another way that I find easier. I'll just copy the book on this one
Equation- 5x^2-13x+6 (same type of equation, just dfferent numbers. you have an x^2, an x and a constant.)
To find the coefficents of the x-terms, you must find the two numbers whose product is 5X6 or 30, and whose sum is -13. The two coefficients must be -10 and -3 since (-10)(-3)=30 and (-10)+(-3)=-13.
Rewrite the expression using -10x and -3x in place of -13x and factor by grouping.
GilesGuthrie
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DoZeRxXx
OK!
By the way:
"As long as algebra is taught in school, there will be prayer in school. " (Cokie Roberts)
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