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So, my problem is 2a^2 + 3a + 1. What I thought I had to do to factor this, would be to figure out what multiplies to equal 2 (2 * 1), and adds to 3. So, I figured it would be 2 and 1, making the answer (2a + 2)(a + 1). But when I checked it, this doesn't work. When I looked in the back of the book to check my answer it was (2a + 1)(a + 1). I'm not sure how I would get this answer because 1 + 1 = 2 and 1 * 1 = 1, which doesn't go along with the method I tried to use originally. Why is this?

- sundaybest
**Posts:**1**Joined:**Thu Jan 03, 2013 11:30 pm

When the coefficient of the x squared term is not 1 you need to use the method of decomposition to factor the trinomial.

The first part you have done correctly.

Find two numbers whose product is a*c and whose sum is b.

In this case a*c=2 and b=3 so the two numbers are 2 and 1.

Now, since the x squared coefficient is not 1 you must use the method of decomposition to factor the trinomial.

You replace the middle term with the two numbers you found. So 2a^2+3a+1 gets rewritten as 2a^2 +2a + a + 1.

Then group the first two terms and last two terms factoring out the largest factor you can out of each group.

So you have 2a(a+1)+(a+1).

Now take (a+1) out as a common factor giving (a+1)(2a+1)

Here is a link with a very clear example of how this works.

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/kim1.html

The first part you have done correctly.

Find two numbers whose product is a*c and whose sum is b.

In this case a*c=2 and b=3 so the two numbers are 2 and 1.

Now, since the x squared coefficient is not 1 you must use the method of decomposition to factor the trinomial.

You replace the middle term with the two numbers you found. So 2a^2+3a+1 gets rewritten as 2a^2 +2a + a + 1.

Then group the first two terms and last two terms factoring out the largest factor you can out of each group.

So you have 2a(a+1)+(a+1).

Now take (a+1) out as a common factor giving (a+1)(2a+1)

Here is a link with a very clear example of how this works.

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/h/kim1.html

- alan
**Posts:**3**Joined:**Thu Jan 03, 2013 11:28 pm

A somewhat less refined but logical approach is to fill this with possibilities to test:

...but will anything work? Your only choices to make product of 1 are 1 X 1.

- Code: Select all
`(2a + ___)(a + ___)`

...but will anything work? Your only choices to make product of 1 are 1 X 1.

- jg.allinsymbols
**Posts:**71**Joined:**Sat Dec 29, 2012 2:42 am

Those are REALLY good suggestions, the last one is actually the fastest way to solve the problem.

But remember that if you ever get stuck, there is a guaranteed solution to any quadratic equation using the Quadratic Formula (and if there isn't, it means the equation is unsolvable, or at least has no real solutions)

x = -b +- SQRT(b^{2} - 4ac]) all-over-2a

But remember that if you ever get stuck, there is a guaranteed solution to any quadratic equation using the Quadratic Formula (and if there isn't, it means the equation is unsolvable, or at least has no real solutions)

x = -b +- SQRT(b

- strumbore
**Posts:**6**Joined:**Fri Jan 11, 2013 12:01 am

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