Question

Factor completely: 3x2 + 5x + 1

(Could also someone explain how to do this? I'm confused... :/ Thank you.)

Answer 1

@mathmale Do you know how to do these?

Answer 2

Looks like this is a `prime polynomial` over integers/rationals

Answer 3

As @ganeshie8 said, you can't factor this. However, there is a polynomial in the neighborhood which can be factored: \[3x^2+5x+2\]and I'll do that as a demonstration of the method.

If you have a trinomial (3 terms) like this, one reliable method for factoring it (if it can be factored) is called factoring by grouping. Here's what you do:

1) first, see if there are any factors common to all 3 terms. If there are, divide them out and set them aside for later. For example, \(6x^2+10x+4\) has a common factor of 2, and can be written as \(2(3x^2+5x+2)\). Getting rid of the \(2\) will make our life a little easier, so just set it aside until you're done with the factoring, then put it back as a multiplier.

2) Now multiply the coefficients of the term with the largest exponent (usually the one in front, but it doesn't have to be) and the constant term (the one which is just a number, usually the last one, but again, it doesn't have to be). The standard form for the polynomial being factored is \[ax^2+bx+c\]with \(a,b,c\) being the coefficients and \(x\) the variable. We are going to multiply \(a*c\).

3) Next, you need to find a pair of factors of \(a*c\) which add up to \(b\). Let's call them \(f_1\) and \(f_2\).

\(f_1 + f_2 = b\) so we can rewrite our polynomial like this:
\[ax^2 + (f_1+f_2)x + c\]Now here's where the magic starts to happen :-)

4) We "group" our terms differently:
\[(ax^2 + f_1x) + (f_2x + c)\]
It doesn't matter which order we choose for \(f_1\) and \(f_2\); we'll end up with the same answer, though the intermediate steps might look a little different. As you practice the method, you should do a few problems where you do it both ways to convince yourself that it doesn't matter.

Now, a safety tip: always write a "+" in front of the second set of parentheses, even if it starts with a term with a negative coefficient! An example of what I mean:
\[(x^2+3x) + (-2x -6)\]rather than\[(x^2+3x)-(2x-6)\](which is wrong) or \[(x^2+3x)-(2x+6)\](which is right). Doing it with my safety tip helps ensure you don't get nailed by a frequent factoring/distribution mistake I see all too often.

5) Factor each group separately. The first group should have a factor of a number and the variable, and the second group will probably have a factor of a number (sometimes it will only be 1 or -1). For the second group, again, if you end up with a negative number as a factor, write it in parentheses. Using my example:
\[(x^2+3x)+(-2x-6)\]\[x(x+3)+(-2)(x+3)\]

6) If everything has gone correctly, you will notice that both halves share a common factor (in parentheses). You'll factor this out and put the two things that multiply that common factor in another set of parentheses:
\[(x+3)(x+(-2))\]\[(x+3)(x-2)\]This is perhaps the most difficult step for newcomers to wrap their heads around, but it makes sense if you've learned how to multiply by the distributive principle instead of that awful FOIL crutch. Looking at it in reverse:
\[(x+3)(x-2) = x(x-2) + 3(x-2)\]by the distributive property
\[x(x-2)+3(x-2) = x*x - 2*x + 3*x -3*2 = x^2-2x+3x-6 \]and you should recognize that as the start of my example, just with the middle two terms switched in order.

7) restore any common factors we set aside at the beginning

8) Finally, it's an excellent idea to multiply your new factoring out and verify that you do get the same polynomial you started with!

Answer 4

So, I'll do that \[6x^2+10x+4\]polynomial as an example for you:

Look for common factors:
\[2(3x^2+5x+2)\]We'll set the \(2\) aside for later and just factor \(3x^2+5x+2\) right now.

Find the \(a*c\) product:
3*2 = 6

Find the factors of the \(a*c\) product that sum to \(b\):
We need two numbers which multiply to 6 and add to 5.

1*6 = 6, 1+6 = 7
2*3 = 6, 2+3 = 5
3*2 = 6, 3+2 = 5
6*1 = 6, 6+1 = 7
-1*-6 = 6, -1+-6 = -7
-2*-3 = 6, -2+-3 = -5
-3*-2 = 6, -3+-2 = -5
-6*-1 = 6, -6+-1 = -7

Clearly, 2 and 3 are the ones we want as \(f_1,f_2\)

Rewrite our middle term
\[3x^2+2x + 3x + 2\]
Group it
\[(3x^2+2x)+(3x+2)\]Factor each group\[x(3x+2)+1(3x+2)\]Factor out the common factor from both groups\[(3x+2)(x+1)\]Restore the common factor we set aside earlier\[2(3x+2)(x+1)\]

Check the result by multiplication:
\[2(3x+2)(x+1) = (6x+4)(x+1) = 6x*x + 6x*1+4*x + 4*1 \]\[\qquad= 6x^2 + 6x+4x + 4\]\[\qquad=6x^2+10x+4\checkmark\]

Thus we have factored \[6x^2+10x+4 = 2(3x+2)(x+1)\]and verified our result.

Answer 5

Thank you so much,, it makes much more sense now.