2m-11m+15. How is this factored?

## How can we factor a trinomials with a prime number

# 2 Answers

Could you possibly have made a mistake and meant to want to know how to factor the following:

2m^{2} - 11m + 15 .....?

If so, then the first step is to factor out the greatest common factor (gcf). In the problem above, there is no gcf.

2m^{2} - 11m + 15 ==> ( )( )

Next, write all of the variables from the first term of the trinomial with half their exponents on the left side of each set of parentheses. That is,

( m )( m )

Then, find all pairs of factors of the coefficients of the first and last terms in the original trinomial.

Coefficient of first term: 2 ==> factors of 2: 1 * 2

Coefficient of last term: 15 ==> factors of 15: 1 * 15 , 3 * 5

Now we choose a pair of factors of 2 and a pair of factors of 15 to insert into the parentheses. Since there is only one pair of factors for 2, we insert a 2 on the left hand side of one of the parantheses and a 1 on the left hand side of the other parentheses. That is,

( 2m )( 1m )

Next, we choose a pair of factors of 15. This will be based of trial and error, but we know that the middle term in the trinomial is negative and the last term is positive, which means that after we use the FOIL method the middle terms must add up to be negative while at the same time the terms on the right hand side of the parentheses must multiply out to give a positive number. Knowing that the only two options you have are to multiply a positive by a positive to get a positive OR multiply a negative by a negative to get a positive, a negative sign must go into both sets of parentheses.

( 2m - )( 1m - )

Let's try the first pair of factors for 15 (i.e., 1 * 15):

( 2m - 1 )( 1m - 15 ) ==> 2m^{2} - 30m - 1m + 15 = 2m^{2} - 31m + 15

This is incorrect.

Let's try switching the places of 1 and 15:

( 2m - 15 )( 1m - 1 ) ==> 2m^{2} - 2m - 15m + 15 = 2m^{2} - 17m + 15

This is incorrect.

We move on to the next pair of factors, 3 * 5 :

( 2m - 3 )( 1m - 5 ) ==> 2m^{2} - 10m - 3m + 15 = 2m^{2} - 13m + 15

This is incorrect.

Switching the placement of the 3 and 15, we get:

( 2m - 5 )( 1m - 3 ) ==> 2m^{2} - 6m - 5m + 15 = 2m^{2} - 11m + 15

This is correct.

Since the result of the last trial is equivalent to the original trinomial, then

2m^{2} - 11m + 15 ==> ( 2m - 5 )( m - 3)

There is a method for solving this problem known as the A-C method of factoring. You can multiply the coefficient of the first term by the coefficient of the last term to find two numbers to split the middle term into.

2m^{2 }- 11m + 15 can be solved as follows:

Step 1:

Multiply the first coefficient by the last (2 * 15 = 30)

Step 2:

Find 2 numbers that multiply to be 30 that add to be -11 (the middle term)

2 numbers that work are -5 and -6

Step 3:

Split the middle term using these two new numbers:

2m^{2} - 5m - 6m + 15

Step 4:

Factor by Grouping

m(2m-5) -3(2m-5)

==> (2m-5)(m-3)

** This method works easily for all factorable trinomials that have a leading coefficient other than 1

## Comments

Nice answer... one minor typo... "Switching the placement of the 3 and 15...".

Also, for benefit of the OP, not all trinomials will factor cleanly. If they do, that's great. If not, you can also use other methods - completing the square, or quadratic equation.

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