Hints for solving the problems(7.1)

Factoring Polynomials I

Greatest Common Factor

Remember monomials are expressions that have one term.

Problem No. 1 - The monomials are 8m³n and 23. 3/x is not a monomial because there is an x in the denominator and x can be zero (0). Expressions that have a zero (0) in the denominator are undefined.

Problem No. 3. - Find the GCF of 12a³b and 16a The GCF is the Greatest Common Factor. The GCF is a the greatest number that will divide evenly into both given numbers. To find the GCF do the following:

         12 = 2 2 3 =    2² 3       prime factors of 12

         16 = 2 2 2     2 = 2 prime factors of 16

                 12a³b    16ab

Use the factors that are common to both numbers, raised to the smallest power shown.

               2 is common to both and the smallest power is squared.

               a is common to both and the smallest power is to the first.

               b is common to both and the smallest power is to the first.

        The GCF is 2²ab  =  4ab

Problem No. 6 - Find the GCF of   9xy²z³, 24xy³z, 18x³yz

      The GCF between   9, 24, & 18   is   3.

      There is an x in every number, the lowest exponent is 1.

      There is a y in each number, the lowest exponent is 1.

      There is a z in each number, the lowest exponent is 3.

        So the GCF is:   3xyz³   The Answer

Problem No. 8 - Factor:  5a³b + 10b   To factor, first look for the GCF and in this problem it is: 5b. To factor out a 5b you must multiply the whole binomial by 5b and divide each term by 5b.

        5b    The problem has the same value, because you are multiplying and dividing the binomial by the same number.

         5b    =    5b(a³ + 2)      The answer

Problem No. 17 - Factor:    

Factor out the GCF first, which is 2mn. Why? 2 is the greatest factor of 4, 10, & 18 and there is an mn in each term. The lowest power shown for m is m and also for n is n.

Factor out the GCF Remember when you factor out the GCF, you multiply the whole expression by the GCF and divide each term by the GCF.

           =     2mn(2 + 5n² - 9m³)      The answer

Problem No. 23 - Factor:      

The GCF is: 3xy²z³ Why? The greatest factor that will go into 9, 15, 21 is 3. There is an xyz in each term and the smallest exponent shown for x is x, for y is y², and for z is z³.

    Factor out the GCF     

                                               3xy²z³(3 - 5x²y³z + 7x³z²)       The answer

Factoring by Grouping

Factoring by grouping is the same as the GCF above, but the GCF is in the form of a binomial in parenthesis, which is a monomial.

Problem No. 30 - Factor: a(b - c) + c(b - c) Notice the expression has two terms. The plus sign separates the terms. Each term has a common factor that is in the form of a binomial, which is (b - c). Factor out the (b - c) and what is left put in a binomial form.

            Multiply and divide the expression by the GCF.

            Notice the (b - c) reduced out in each term leaving.

                (b - c)(a + c)         The answer

Problem No. 36 - Factor: a(3a - b) - b(3a - b) The GCF in the form of a binomial is: (3a - b). Put the (3a - b) on the outside and divide every term of the problem by it and your answer is:

                     (3a - b)(a - b)

Problem No. 40 - Factor: mp - mq + np - nq This problem is a little different as it doesn’t show the GCE as a binomial yet. Most of the time, when you have 4 terms to factor, you must group first. Usually you will group the first two and the second two.

           (mp - mq) + (np - nq)       Now factor the GCF of each binomial

            m(p - q)  +   n(p - q)       Oh, wow, we not have two terms with a GCF in the form of a binomial. The binomial in the parenthesis must be the same for both terms. If it isn’t, then start over with the grouping. Factor it out, just as you did above.

                  (p - q)(m + n)     The answer

Problem No. 41 - Factor: ac + ad - bc - bd Four terms, group the first two and the second two. But the first two have a plus sign between them and the second two have a minus between. So when you group the first two put the plus sign between and when you group the second two put the negative (minus) sign in front and a plus sign between the terms. When you group terms and put a negative sign in the front, you have to change the sign of each term that follows it. The bc and bd had a negative sign in the front of it, so when you group it, it looks like -(bc + bd). Why do this? Because

the binomials have to have the same sign between them.

              (ac + ad) - (bc + bd)      Grouping

                a(c + d) - b(c + d) Factoring the GCF and same binomial in each term is left.

                    (c + d)(a - b) Factoring out the (c + d) and the answer.

Problem No. 55 - Factor: 12pr² - 16rs - 20s + 15pr

Four terms, group the first two and the last two will not produce a GCF in the form of the same binomial. Try it and see.

So let’s group the first and fourth term together and the second and third term together.

(12pr² + 15pr) - (16rs + 20s) Remember when you put a negative sign in front of a parenthesis, you must change all the signs of the terms in the parenthesis from their original. Does that make sense?

      3pr(4r + 5) - 4s(4r + 5)     Same GCF as a binomial in each term.

        (4r + 5)(3pr - 4s) Factoring out the GCF as a binomial and that is the answer.

The binomial answers above can be multiplied together to get the original four term expression.