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The common term for that phrase is the verb 'factor'. Factoring is the reverse of expanding (aka: multiplying out, or 'foiling, in this instance').X^2 + 2X + 2

If you

\(\displaystyle \text{reverse f}\text{oil}\)

… would you get:

(X+1)(X+1)

? Just a guess.

X^2+X+X+1

Nope, that doesn't work.

The given quadratic polynomial does not factor nicely. The factorization is:

(x + 1 -

where symbol

There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely. :cool:

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Thanks for answering.The common term for that phrase is the verb 'factor'. Factoring is the reverse of expanding (aka foiling, in this instance).

The given quadratic polynomial does not factor nicely. The factorization is:

(x + 1 -)(x + 1 +i)i

where symbolrepresents the square root of -1 (aka the imaginary unit).i

There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely. :cool:

Is there a way to find the lowest common denominator to factor?

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I'm not sure that I understand. There's no denominator, in your example.… Is there a way to find the lowest common denominator to factor?

OK, I will try and come up with an example.I'm not sure that I understand. There's no denominator, in your example.

X^2+40X+6

X^2, 40, and 6 all have the number 2 in common. X^2 is just X times X, so you really don't need to manipulate anything there.

Could you use 2 to find the lowest common denominator to factor the rest of the equation out?

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You may mean "greatest common factor" rather than "lowest common denominator".OK, I will try and come up with an example.

X^2+40X+6

X^2, 40, and 6 all have the number 2 in common. X^2 is just X times X, so you really don't need to manipulate anything there.

Could you use 2 to find the lowest common denominator to factor the rest of the equation out?

But 2 is

This polynomial, like the first, can't be factored over the integers (that is, factored into factors containing only integers).

Thank you for your answer. It answers my question.You may mean "greatest common factor" rather than "lowest common denominator".

But 2 isnota factor of x^2; it's an exponent, which is an entirely different thing.

This polynomial, like the first, can't be factored over the integers (that is, factored into factors containing only integers).

Thanks, I think I have the basic concept down now.You can look at it this way:

your math teacher makes up a similar problem:

(2x + 3) * (x + 7) = 2x^2 + 17x + 21

Then tells you: factor 2x^2 + 17x + 21

Your answer should be (2x + 3)(x + 7)

You have to take the equation as a whole and you can't split it up into different sections. That is what I was trying to figure out.

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Right; that's useless for your goal.But I was thinking along these lines:

X^2+40X+6

x^2+2(20x+3)

But I don't think this actually changes the equation to the point that you can Factor from that point...

As I said previously, "This polynomial, like the first,

The fact is, factoring is not possible in general. If you make up a polynomial randomly, it will most likely be unfactorable (in the sense you are asking for).

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X^2 + 40X + 6There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely.

A = 1

B = 40

C = 6

Discriminant = 40^2 - 4(1)(6) = 1576

sqrt(1576) ≈ 39.6989

The Discriminant is not a perfect square. Hence, the given polynomial does not factor nicely. :cool:

Thanks for answering.Right; that's useless for your goal.

As I said previously, "This polynomial, like the first,can'tbe factored over the integers (that is, factored into factors containing only integers)." In other words, you can't do what you want to do. More than that,Ican't do what you want to do -- no one can do it, unless you expand the goal to allow irrational numbers.

The fact is, factoring is not possible in general. If you make up a polynomial randomly, it will most likely be unfactorable (in the sense you are asking for).

I would just like to add that in this case, though it doesn't forward my goal to try and simplify part of the equation to solve the equation, it is my goal to learn, and that is what I have done in this thread.

I know I am maybe a bit toooo experimental in my approach in learning what works and what doesn't, but this is the way I learn best...

I had a goal to learn math from Khan Academy (and it's still a valuable recourse if I have something specific I want to look up), but if I can keep asking questions here, I think this is a way to learn that suits me a lot better. I would like to be at the point where I know basic algebra fairly well by the time I enroll in college, and I think I am on that pace so far. So that said, I really do appreciate the mods here - especially considering they are all volunteers.

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Quick note: you've been confusing "mods" with "volunteer tutors".… I really do appreciate the mods here …

We have well over a dozen tutors who volunteer regularly. The forum currently has four active moderators. The moderators handle housekeeping tasks and policy issues, in addition to tutoring. :cool:

Oh. Thanks for explaining that. It makes more sense.Quick note: you've been confusing "mods" with "volunteer tutors".

We have well over a dozen tutors who volunteer regularly. The forum currently has four active moderators. The moderators handle housekeeping tasks and policy issues, in addition to tutoring. :cool:

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I'm not sure what you mean by "splitting into different sections"...? Also, a quadratic, such as what you've posted, is not an "equation". An equation is a statement of equality; thus, it has an "equals" sign (and, possibly, can be solved). What you have posted are "expressions", which can be "simplified", "factored", etc.Thanks, I think I have the basic concept down now.

You have to take the equation as a whole and you can't split it up into different sections. That is what I was trying to figure out.

To learn about finding the greatest common factor in polynomial expressions, please try

I see.X^2 + 40X + 6

A = 1

B = 40

C = 6

Discriminant = 40^2 - 4(1)(6) = 1576

sqrt(1576) ≈ 39.6989

The Discriminant is not a perfect square. Hence, the given polynomial does not factor nicely. :cool:

how do you get the 4 in there?

As I understand it if you have anything that could equal a perfect square from the formula above then it it factorable correct?

So if you got something that ended up equaling 40^2 (or any number squared) then it is factorable. I assume it's required that B squared has to be low enough for the remaining -4(a)(c) to get a sum to another higher square, right? So then this would mean that B has to be lower than C unless you are dealing with negatives, right?

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The factor 4 is a part of the formula for the Discriminant.how do you get the 4 in there?

There's a shortcut for determining whether a quadratic polynomial (Ax^2+Bx+C) factors nicely. Determine its Discriminant (B^2-4AC). If the Discriminant is not a perfect square, then the polynomial does not factor nicely.

If the value of the Discriminant is a perfect square, then, yes, the quadratic polynomial will factorAs I understand it if you have anything that could equal a perfect square from the formula above then it it factorable correct?

If you gotSo if you gotsomethingthat ended up equaling …any numbersquared … then it isfactorable.

For me, a nice factorization doesn't necessarily mean that only Integers appear with the factors. There can be nice factorizations involving Rational numbers. (Plus, the definition of a perfect square IS the square of a Rational number.) Here's a quick example:

x^2 + x/6 - 1/3 = (x - 1/2)(x + 2/3)

The Discriminant for this quadratic polynomial is 49/36. That's a perfect square.

I've got to leave for dinner. I will think about this, when I return. Cheers :cool:I assume it's required that B squared has to be low enough for the remaining -4(a)(c) to get a sum to another higher square, right? So then this would mean that B has to be lower than C unless you are dealing with negatives, right?

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You can, however, deduce a few important things about your factorization by looking at the equation and thinking about it logically, to reduce the work to find the answer. Let's reverse engineer 2x^2 + 17x + 21, since it was provided above.Thanks, I think I have the basic concept down now.

You have to take the equation as a whole and you can't split it up into different sections. That is what I was trying to figure out.

First obvious thing is x^2. It means we're going to have an X on the same side of both equations. (x...something)(x...something)

Then we look at the signs in the equation above. + and +. The ONLY way we're going to see this is if both equations have + operators in them. (x + something)(x + something)

Now we look at the 2 in the

All that's left at this point is to sort out how we can get +21. The possible factors here are (1 and 21) and (3 and 7). Since we know we're only adding, we can't add anything more to 21 to get a sum of 17, so we can discard it as a possible solution, which means 3 and 7 have to be the factors. We just need to try and place them with the 2 and 1 so that they add up to 17.

(2x + 7)(x + 3).. if we just mentally plug them in like this, we can quickly see that 6x + 7x isn't going to give us 17x... All we have to do is swap them and then try again.

(2x + 3)(x + 7).. 14x + 3x = 17x...

PING! We've deduced the solution with just a little logical thinking and very little hard math-stuff needed at all.

Just break it down step by step and you can often find the solution to these type problems in your head, without even needing a piece of paper to scribble notes down on. (Not that your teacher will approve of you just writing the answer, if you calculate it mentally, as that makes it hard to complete the "show your work" requirement.)