Now write the common factor, 4a1/2, outside the parentheses and the results of the division inside.
5. Factor out the GCF: 24x2y3 – 42x3y2.
Solve It
6. Factor out the GCF: .
Solve It
7. Factor out the GCF: 16a2b3c4 – 48ab4c2.
Solve It
8. Factor out the GCF: .
Solve It
Reducing Algebraic Fractions
The basic principles behind reducing fractions with numbers and reducing fractions with variables and numbers remain the same. You want to find something that divides both the numerator (the top of the fraction) and denominator (bottom of the fraction) evenly and then leave the results of the division as the new numerator and new denominator.
When the fraction has two or more terms in the numerator, denominator, or both, you first have to factor out the GCF before you can reduce. And when the algebraic fraction has just multiplication and division in the numerator and denominator, the reducing part is pretty easy. Just divide out the common factors as shown in the following example.
Q. Reduce the fraction: .
A.
In this fraction, the GCF is 7xy:
Q. Reduce the fraction: .
A.
First find the GCF of the numerator and denominator; it’s :
9. Reduce the fraction: .
Solve It
10. Reduce the fraction: .
Solve It
11. Reduce the fraction: .
Solve It
12. Reduce the fraction: .
Solve It
13. Reduce the fraction: .
Solve It
14. Reduce the fraction: .
Solve It
Answers to Problems on Factoring Expressions
This section provides the answers (in bold) to the practice problems in this chapter.
1. Write the prime factorization of 24. The answer is .
2. Write the prime factorization of 100. The answer is .
3. Write the prime factorization of 256. The answer is 28.
4. Write the prime factorization of 3,872. The answer is .
5. Factor out the GCF: .The answer is 6x2y2(4y – 7x).
6. Factor out the GCF: . The answer is .
When factoring negative exponents, you factor out the smallest power, which is the most negative power, or the number farthest to the left on the number line. In this case, –4 is the smallest power. Notice that the powers in the parentheses are all non-negative.
7. Factor out the GCF: . The answer is .
8. Factor out the GCF: . The answer is . The greatest common factor includes the negative exponent on the x factor. Remember that –4 is smaller than –3; subtract the exponents when dividing.
9. Reduce the fraction: . The answer is . All three terms have the common factor (x + 2).
10. Reduce the fraction: . The answer is 30a7b.
For a refresher on factorials, refer to Chapter 6.
11. Reduce the fraction: . The answer is .
12. Reduce the fraction: . The answer is .
Even though the answer appears to have a common factor in the numerator and denominator, you can’t reduce it. The numerator has two terms, and the first term, the 3, doesn’t have that common factor in it.
13. Reduce the fraction: . The answer is .
When factoring the terms in the numerator, be careful with the subtraction of the fractions. These fractional exponents are found frequently in higher mathematics and behave just as you see here. I put the exponent of 1 on the x in the numerator just to emphasize the result of the subtraction of exponents. Continuing,
14. Reduce the fraction: . The answer is .
Factor the numerator and denominator separately and then reduce by dividing by the common factors in each:
Chapter 10
Taking the Bite Out of Binomial Factoring
In This Chapter
Making quick work of the difference between two squares
Dicing up the cubes — their sums and differences
Using multiple factoring techniques
You have several different choices when factoring a binomial (an expression that is the sum or difference of two terms):
Factor out a greatest common factor (GCF).
Write the expression as the product of two binomials — one the sum of the two roots and the other the difference of those same two roots.
Write the expression as the product of a binomial and trinomial (an expression with three terms) — one with the sum or difference of the cube roots and the other with squares of the roots and a product of the roots.
Use two or more of the above.
In this chapter, you find how to recognize which type of factorization to use and how to change from two terms to one by factoring the expression. I cover GCF in Chapter 9, if you need a refresher. The other procedures are new to this chapter.
Factoring the Difference of Squares
When a binomial is the difference of two perfect squares, you can factor it into the product of the sum and difference of the square roots of those two terms:
a2 – b2 = (a + b)(a – b)
Q. Factor: 4x2 – 81.
A. (2x + 9)(2x – 9). The square root of 4x2 is 2x, and the square root of 81 is 9.
Q. Factor: 25 – 36x4y2z6.
A. (5 + 6x2yz3)(5 – 6x2yz3). The square of 6x2yz3 is 36x4y2z6. Notice that each exponent is doubled in the square.
1. Factor: x2 – 25.
Solve It
2. Factor: 64a2 – y2.
Solve It
3. Factor: 49x2y2 – 9z2w4.
Solve It
4. Factor: .
Solve It
Factoring Differences and Sums of Cubes
When you have two perfect squares, you can use the special factoring rule if the operation is subtraction. With cubes, though, both sums and differences factor into the product of a binomial and a trinomial.
a3 – b3 = (a – b)(a2 + ab + b2) and a3 + b3 = (a + b)(a2 – ab + b2)
Here’s the pattern: First, you write the sum or difference of the two cube roots corresponding to the sum or difference of cubes; second, you multiply the binomial containing the roots by a trinomial composed of the squares of those two cube roots and the opposite of the product of them. If the binomial has a + sign, the middle term of the trinomial is –. If the binomial has a – sign, then the middle term in the trinomial is +. The two squares in the trinomial are always positive.
Q. Factor: x3 – 27.
A. (x – 3)(x2 + 3x + 9)
x3 – 27 = (x – 3)(x2 + x · 3 + 32) = (x – 3)(x2 + 3x +9)
Q. Factor: 125 + 8y3.
A. (5 + 2y)(25 – 10y + 4y2)
125 + 8y3 = (5 + 2y)(52 – 5 · 2y +[2y]2) = (5 + 2y)(25 – 10y + 4y2)
5. Factor: x3 + 1.
Solve It
6. Factor: 8 – y3.
Solve It
7. Factor: 27z3 + 125.
Solve It
8. Factor: 64x3 – 343y6.
Solve It
Making Factoring a Multiple Mission
Many factorization problems in mathematics involve more than one type of factoring process. You may find a GCF in the terms, and then you may recognize that what’s left is the difference of two cubes. You sometimes factor the difference of two squares just to find that one of those binomials is the difference of two new squares.
Solving these problems is really like figuring out a gigantic puzzle. You discover how to conquer it by applying the factorization rules. In general, first look for a GCF. Life is much easier when the numbers and powers are smaller because they’re easier to deal with and work out in your head.
Q. Factor: 4x6 + 108x3.
A. 4x3 (x + 3)(x2 – 3x + 9). First, take out the GCF, 4x3. Then factor the sum of the cubes in the parentheses:
4x6 + 108x3 = 4x3(x3 + 27) = 4x3(x + 3)(x2 – 3x + 9)
Q. Factor: y8 – 256.
A. (y4 + 16)(y2 + 4)(y + 2)(y – 2
). You can factor this problem as the difference of two squares. Then the second factor factors again and again:
y8 – 256 = (y4 +16)(y4 – 16) = (y4 + 16)(y2 + 4)(y2 – 4) = (y4 + 16)(y2 + 4)(y + 2)(y – 2)
9. Completely factor: 3x3y3 – 27xy3.
Solve It
10. Completely factor: 36x2 – 100y2.
Solve It
11. Completely factor: 80y4 – 10y.
Solve It
12. Completely factor: 10,000x4 – 1.
Solve It
13. Completely factor: x–4 + x–7.
Solve It
14. Completely factor: 125a3b3 – 125c6.
Solve It
Answers to Problems on Factoring
This section provides the answers (in bold) to the practice problems in this chapter.
1. Factor: . The answer is .
2. Factor: .The answer is .
3. Factor: . The answer is .
4. Factor: . The answer is .
Whenlooking at the exponents,you see that 1⁄2 is twice the fraction 1⁄4, and 1⁄4 is twice the fraction 1⁄8.
5. Factor: . The answer is .
6. Factor: . The answer is .
It’s nice to have a list of the first ten cubes handy when factoring the sum or difference of cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
7. Factor: . The answer is .
8. Factor: . The answer is .
Did you remember that ?
9. Completely factor: . The answer is .
10. Completely factor: . The answer is .
11. Completely factor: . The answer is .
12. Completely factor: . The answer is .
13. Completely factor: . The answer is .
You first take out the GCF, which involves the most negative exponent. The resulting binomial in the parentheses is the sum of two perfect cubes.
14. Completely factor: . The answer is .
You may have been tempted to go right into the difference of cubes because 125 is a perfect cube. It’s always more desirable, though, to factor out large numbers when possible.
Chapter 11
Factoring Trinomials and Special Polynomials
In This Chapter
Zeroing in on the greatest common factor
Reversing FOIL for a factorization
Assigning terms to groups to factor them
Making use of multiple methods to factor
In Chapter 10, you find the basic ways to factor a binomial (an expression with two terms). Factoring means to change the expression from several terms to one expression connected by multiplication and division. When dealing with a polynomial with four terms, such as x4 – 4x3 – 11x2 – 6x, the four terms become one when you write the factored form using multiplication: x(x + 1)2(x – 6). The factored form has many advantages, especially when you want to simplify fractions, solve equations, or graph functions.
When working with an algebraic expression with three terms (a trinomial) or more terms, you have a number of different methods available for factoring it. You generally start with the greatest common factor (GCF) and then apply one or more of the other techniques if necessary. This chapter covers the different methods and provides several sample questions for you to try.
Focusing First on the Greatest Common Factor (GCF)
In any factoring problem, first you want to find a common factor — if one exists. If you find a GCF for the terms in the expression, then you divide every term by that common factor and write the expression as the product of the common factor and the results of each division.
Q. Factor out the GCF: 28x2y – 21x3y2 + 35x5y3.
A. 7x2y(4 – 3xy + 5x3y2)
You divide every term by the GCF, 7x2y.
Q. Factor out the binomial GCF: 3(x – 5)4 + 2a(x – 5)3 – 11a2(x – 5)2.
A. (x – 5)2[3(x – 5)2 + 2a(x – 5) – 11a2] The GCF is the square of the binomial, (x – 5)2. This example has three terms with a varying number of factors, or powers, of the binomial (x – 5) in each term.
1. Factor out the GCF: 8x3y2 – 4x2y3 + 14xy4.
Solve It
2. Factor out the GCF: 36w4 – 24w3 – 48w2.
Solve It
3. Factor out the GCF: 15(x – 3)3 + 60x4(x – 3)2 + 5(x – 3).
Solve It
4. Factor out the GCF: 5abcd + 10a2bcd + 30bcde + 20b3c2d.
Solve It
“Un”wrapping the FOIL
The FOIL method helps you when you’re multiplying two binomials. (See Chapter 7 for some problems that use this process.) When factoring, you unFOIL, which reverses the process to tell you which two binomials were multiplied together in the first place. The task in this chapter is to first recognize that a trinomial has been created by multiplying two binomials together and then factor it — figure out what those binomials are and write the product of them.
The general procedure for performing unFOIL includes these steps:
1. Write the trinomial in descending powers of a variable.
2. Find all the possible combinations of factors whose product gives you the first term in the trinomial.
3. Find all the possible sets of factors whose product gives you the last term in the trinomial.
4. Try different combinations of those choices of factors in the binomials so that the middle term is the result of combining the outer and inner products.
Q. Factor 2x2 – 5x – 3.
A. (2x + 1)(x – 3)
1. The trinomial is already written with descending powers.
2. The only possible factors for 2x2 are 2x and x.
3. The only possible factors for the last term are 3 and 1.
4. Work on creating the middle term. Because the last term is negative, you want to find a way to arrange the factors so that the outer and inner products have a difference of 5x. You do this by placing the 2x and 3 so they multiply one another: (2x 1)(x 3). When deciding on the placement of the signs, use a + and a –, and situate them so that the middle term is negative. Putting the – sign in front of the 3 results in a –6x and a +1x. Combining them gives you the –5x.
Q. Factor 12y2 – 17y + 6.
A. (4y – 3)(3y – 2). The factors of the first term are either y and 12y, 2y and 6y, or 3y and 4y. The factors of the last term are either 1 and 6 or 2 and 3. The last term is positive, so the outer and inner products have to have a sum of 17y. The signs between the terms are negative (two positives multiplied together give you a positive) because the sign of the middle term in the original problem is negative.
5. Factor x2 – 8x + 15.
Solve It
Algebra I Workbook For Dummies Page 9