• Look for the greatest factor common to every term

2

• Factor out the GCF by dividing it into each term

• In this problem, the greatest common factor includes both numbers and variables. First we need to factor out the greatest number that will divide into both 15 and 9. In this case, it will be a 3. Next we need to factor out the smallest power of the variable x that can be seen in the problem. In this case, it will be . Our GCF is

• Factor out the GCF by dividing it into each term.

• Find the greatest factor common to every term. Since the last term, 24, does not contain any variables, no variables are “common” in this problem and only the -6 may be factored out.

-6

• Factor out the GCF by dividing it into each term

• Factor out 3a from the first 2 terms and 4 from the last 2 terms.

• Notice that the terms inside each set of parentheses are the same. Those terms have now become the GCF. The answer may be checked by multiplying the factored form back out to see if you get the original polynomial.

• Factor out 2a from the first 2 terms and -5 from the last 2 terms. Be careful about signs!

• The terms inside each set of parentheses are the same. Those terms have now become the GCF.

• Examine the problem for a GCF. There is none.

• To factor a difference in two squares, use two sets of parentheses.

• Take the square root of each term. The square root of a variable’s exponent will be half of the exponent.

and

• Use the square roots to fill in the parentheses. Be sure to check that neither factor will factor again. What is the final answer?

• Factor out the GCF of .

• Now factor the difference in two squares. The square root of is and the square root of is . Make certain that you check to be sure that neither factor will factor again. What is the final answer?

• Factor the difference in two squares. The square root of is and the square root of 16 is 4.

• Check to be sure that neither factor will factor again. The term is the difference of two squares again and must be factored again. Notice that the second term is a sum of two squares, and will not factor.

• This is a difference in two cubes, so begin with two sets of parentheses.

• In the first set, there will be a binomial containing the cube root of each term. In this problem, x and 3.

• In the second set there will be a trinomial. The first term of the trinomial is the square of the first term in the binomial.

• The last term is the square of the last term in the binomial.

• The middle term is the product of the two terms in the binomial.

• You will be finished when you insert the appropriate sign between each of the terms.

• This is a sum of two cubes, so begin with two sets of parentheses.

• In the first set, there will be a binomial containing the cube root of each term. In this problem, and .

• In the second set there will be a trinomial. The first term of the trinomial is the square of the first term in the binomial.

• The last term is the square of the last term in the binomial.

• The middle term is the product of the two terms in the binomial.

• You will be finished when you insert the appropriate sign between each of the terms.

• Factor out the GCF first.

• Now finish the problem by factoring the difference of the two perfect cubes.

• Factor out the GCF first.

• Now finish by factoring the sum of the two perfect cubes.

• The three terms are arranged in descending order. There is not a GCF. Therefore the factoring process is begun by opening two sets of parentheses.

• Place the factors for the first term of the trinomial in the front of each set of parentheses.

• Then, because the sign of the last term is positive, factor the last term of the trinomial to factors that multiply to give 12 and add to give 7.

• Finally, because the sign of the last term is positive, the sign of the 4 and the sign of the 3 will each have the same sign. Because the sign of the 7 is positive, the sign of the 4 and the sign of the 3 will each be a positive sign. Check the answer using multiplication.

• This example is very similar to Example #1 above. So we begin by opening two sets of parentheses and placing the factors for the first term in the front of each set of parentheses.

• The difference lies in the signs. The sign of the 12 is still positive,  so the sign of the 4 and the 3 will again be the same. However, since the sign of the 7 is negative, the sign of the 4 and the 3 will each be negative.

• The terms are arranged in descending order, and there is no GCF. So again begin by opening two sets of parentheses and placing factors for the first term in the front of each set of parentheses.

• Choose numbers that multiply to give 12. Since the 12 is negative, find two numbers that subtract to give 4.

• Since the sign of the 12 is negative, one factor of the answer will be positive and the other will be negative. Since the sign of the middle term of the trinomial is negative, the larger of the two factors used in the answer will have that sign. Therefore, the 6 will be negative and the 2 will be positive.

• Factor out the GCF.

• In factoring the general trinomial, begin with the factors of 12. These include the following: 1, 12, 2, 6, 3, 4. As a general rule, the set of factors closest together on a number line should be tried first as possible factors for the trinomial.

• The only factors of the last term of the trinomial are 1 and 3, so there are not other choices to try. Because the last term is negative the signs of the factors 1 and 3 must be opposite.

• This is the first trial.  The answer must be checked by multiplication, as follows:

• The factorization of the trinomial is almost correct. However, the sign of the middle term is incorrect. That means that the signs of the two factors should be switched.

• Factor out the negative sign first. Doing so will change all the signs of the trinomial.

• Now factor the trinomial. Factors of the first term include 1, 4, 2. Factors of the last term include 1, 6, 2, 3. The sign of the 6 is negative, so the signs in the two factors must be opposites.

• Consider 2 and 2 as factors of 4, and 3 and 2 as factors of 6.

• Such choices are not good, because it causes the second factor to contain a GCF and  that should be avoided. A second  attempt must be made, since checking  the factors will fail as follows:

• Try another combination.

• Check the factorization.

• Trials would continue by perhaps trying to switch the 3 and 2; however, that would cause a GCF in the first set of parentheses. That should be avoided, so the next idea would be to use 6 and 1 instead of 3 and 2.