First we must note that a common factor does not need to be a single term.

For instance, in the expression 2y(x 3) 5(x 3) we have two terms. In each of these terms we have a factor (x 3) that is made up of terms. Solution First note that not all four terms in the expression have a common factor, but that some of them do.

The more you practice this process, the better you will be at factoring.

Now that we have established the pattern of multiplying two binomials, we are ready to factor trinomials.

The expression is now 3(ax 2y) a(ax 2y), and we have a common factor of (ax 2y) and can factor as (ax 2y)(3 a).

Multiplying (ax 2y)(3 a), we get the original expression 3ax 6y a Factor 3ax 2y 3ay 2x.Knowing that the product of two negative numbers is positive, but the sum of two negative numbers is negative, we obtain Solution We are here faced with a negative number for the third term, and this makes the task slightly more difficult.Since -24 can only be the product of a positive number and a negative number, and since the middle term must come from the sum of these numbers, we must think in terms of a difference.In earlier chapters the distinction between terms and factors has been stressed. Note in these examples that we must always regard the entire expression.You should remember that terms are added or subtracted and factors are multiplied. Factors can be made up of terms and terms can contain factors, but factored form must conform to the definition above.We now wish to look at the special case of multiplying two binomials and develop a pattern for this type of multiplication.Since this type of multiplication is so common, it is helpful to be able to find the answer without going through so many steps. From the example (2x 3)(3x - 4) = 6x Not only should this pattern be memorized, but the student should also learn to go from problem to answer without any written steps.A large number of future problems will involve factoring trinomials as products of two binomials.In the previous chapter you learned how to multiply polynomials.Solution Here the problem is only slightly different.We must find numbers that multiply to give 24 and at the same time add to give - 11. The last term is obtained strictly by multiplying, but the middle term comes finally from a sum.

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