Algebra I Workbook For Dummies
Page 10
6. Factor y2 – 6y – 40.
Solve It
7. Factor 2x2 + 3x – 2.
Solve It
8. Factor 4z2 + 12z + 9.
Solve It
9. Factor w2 – 16.
Solve It
10. Factor 12x2 – 8x – 15.
Solve It
Factoring Quadratic-Like Trinomials
A quadratic-like trinomial has a first term whose power on the variable is twice that in the second term. The last term is a constant. In general, these trinomials are of the form . If these trinomials factor, then the factorizations look like . Notice that the power on the variables in the factored form matches the power of the middle term in the original trinomial.
To factor a quadratic-like trinomial, you treat it as if it were ax2 + bx + c, with the same rules applying to unFOILing and just using the higher powers on the variables.
Q. Factor .
A. . Treat this problem as if it is the trinomial . You have to find factors for the first term, involving the 6, and factors for the last term, involving the 28. The middle term, with the 13, has to be the difference between the outer and inner products. Using 3x2 from the first term and 7 from the last term, you get a product of 21x2. Then, using 2x2 from the first and 4 from the last, you get 8x2. The difference between 21 and 8 is 13. Get the factors aligned correctly and the signs inserted in the right places.
Q. Factor .
A. . Don’t let the negative exponents throw you. The middle term has an exponent that’s half the first term’s exponent. Think of the trinomial as being like . It works!
11. Factor x10 + 4x5 + 3.
Solve It
12. Factor 4y16 – 9.
Solve It
13. Factor .
Solve It
14. Factor .
Solve It
Factoring Trinomials Using More Than One Method
You can factor trinomials by taking out a GCF or by using the unFOIL method, and sometimes you can use both of these methods in one expression. Even better, you often get to apply the rules for factoring binomials on one or more of the factors you get after unFOILing. In general, first take out the common factor to make the terms in the expression simpler and the numbers smaller. Then look for patterns in the trinomial that results.
Q. Factor 9x4 – 18x3 – 72x2.
A. 9x2(x – 4)(x + 2). First take out the common factor, 9x2, and write the product with that common factor outside the parentheses: 9x2(x2 – 2x – 8). You can then factor the trinomial inside the parentheses.
Q. Factor .
A. . First take out the common factor, 3x, and write the product with that common factor outside the parentheses: 3x (x4 – 5x2 + 4). You can then factor the quadratic-like trinomial inside the parentheses to get . The two binomials then each factor into the difference of squares.
Q. Factor (x – 3)3 + (x – 3)2 – 30(x – 3).
A. (x – 3)[(x – 8)(x + 3)]. Remember, the GCF can be a binomial. First, factor out the binomial (x – 3), which gives you (x – 3)[(x – 3)2 + (x – 3) – 30]. Then you have two options. This example uses the first option (the next example has the other option): You can square the first term in the brackets and then simplify the expression that results by combining like terms: (x – 3)[x2 – 6x + 9 + x – 3 – 30] = (x – 3)[x2 – 5x – 24]. Now you can factor the trinomial in the brackets.
Q. Factor (x – 3)3 + (x – 3)2 – 30(x – 3) using a quadratic-like pattern.
A. (x – 3)[(x – 8)(x + 3)]. Yes, this is the same problem and same answer as shown in the preceding example. This time, after factoring out the binomial, you use the quadratic-like pattern to factor the expression inside the bracket. Beginning with (x – 3)[(x – 3)2 + (x – 3) – 30], think of the (x – 3) expressions as being y’s, and you see y2 + y – 30, which is a trinomial that factors into (y + 6)(y – 5). Replace the y’s with the binomials, and you have (x – 3 + 6)(x – 3 – 5), which simplifies into (x + 3)(x – 8). You get the same answer.
15. Completely factor 3z2 – 12z + 12.
Solve It
16. Completely factor 5y3 – 5y2 – 10y.
Solve It
17. Completely factor x6 – 18x5 + 81x4.
Solve It
18. Completely factor w4 – 10w2 + 9.
Solve It
19. Completely factor 3x2(x – 2)2 + 9x(x – 2)2 – 12(x – 2)2.
Solve It
20. Completely factor .
Solve It
21. Completely factor .
Solve It
22. Completely factor .
Solve It
Factoring by Grouping
The expression 2axy + 8x – 3ay – 12 has four terms — terms that don’t share a single common factor. But you notice that the first two terms have a common factor of 2x, and the last two terms have a common factor of –3. What to do, what to do!
Don’t worry. This problem suggests that factoring by grouping may be an option. Expressions that can be factored by grouping have two distinct characteristics:
A common factor (or factorable expression) occurs in each pairing or grouping of terms.
The factorization of each individual grouping results in a new GCF common to each group.
Demonstrating this process with some examples makes it easier to understand:
Q. Factor by grouping: 2axy + 8x – 3ay – 12.
A. (ay + 4)(2x – 3). Factor 2x out of the first two terms and –3 out of the second two terms: 2x(ay + 4) – 3(ay + 4). You can see how you now have two terms, instead of four, and the two new terms have a common factor of (ay + 4). If you factor out a 3 instead of a –3, the second term becomes 3(–ay – 4), which doesn’t have the same (ay + 4) as the first. For factoring by grouping to work, the two new common factors have to be exactly the same.
Q. Factor by grouping: 2a3x – a3y – 6b2x + 3b2y + 2cx – cy.
A. (2x – y)(a3 – 3b2 + c). This type of factoring by grouping can also work with six terms. First, find the common factor in each pair of terms. Factor a3 out of the first two terms, – 3b2 out of the second two terms, and c out of the last two: a3(2x – y) – 3b2(2x – y) + c(2x – y). Notice that, with the middle pairing, if you factor out 3b2 instead of –3b2, you don’t have the same common factor as the other two pairings: the signs are wrong. The new trinomial doesn’t factor, but you should always check in case another factorization possibility is lurking in the background.
23. Factor by grouping: ab2 + 2ab + b + 2.
Solve It
24. Factor by grouping: xz2 – 5z2 + 3x – 15.
Solve It
25. Factor by grouping: .
Solve It
26. Factor by grouping: .
Solve It
27. Factor by grouping: ax – 3x + ay – 3y + az – 3z.
Solve It
28. Factor by grouping: x2y2 + 3y2 + x2y + 3y – 6x2 – 18.
Solve It
Putting All the Factoring Together
Factoring an algebraic expression is somewhat like buying presents when the holidays roll around. When you buy presents, you categorize (family, friend, or acquaintances?) and purchase items based on those categories so that the end result — complete satisfaction — doesn’t break the bank. In terms of factoring algebraic expressions, you categorize (is an expression two terms, three terms, or more terms?) and then choose a procedure that works for that category. If you have two terms, for example, you may get to factor them as the difference of squares or cubes or the sum of cubes. If you have three terms, you may be able to factor the expression as the product of two binomials. With four or more terms, grouping may work. In any case, you first want to look for the GCF. Your goal? A completely satisfactory result that doesn’t use too much brain power.
Q. Completely factor 8x3 + 56x2 – 240x.
A. 8x(x + 10)(x –3). The first thing you always look for is a common factor — in this case, it’s 8x — that you can factor out: 8x(x2 + 7x – 30). Then you can use the unFOIL method o
n the trinomial in the parentheses. Finally, write the answer as the factor 8x times the product of the two binomials from the trinomial.
Q. Completely factor 3x5 – 75x3 + 24x2 – 600.
A. 3[(x + 5)(x – 5)(x + 2)(x2 – 2x + 4)]
1. Factor out the common factor of 3.
3(x5 – 25x3 + 8x2 – 200)
2. Factor by grouping.
Take x3 out of the first two terms and 8 out of the last two terms. Then factor out the common factor of the two terms, the binomial.
3[x3(x2 – 25) + 8(x2 – 25)] 3[(x2 – 25)(x3 + 8)]
The first factor in the brackets is the difference of perfect squares. The second factor is the sum of perfect cubes.
3. Factor each of the two binomials in the parentheses to find the answer.
(x2 – 25) = (x – 5)(x + 5)
(x3 + 8) = (x + 2)(x2 – 2x + 4)
3[(x2 – 25)(x3 + 8)] = 3[(x – 5)(x + 5)(x + 2)(x2 – 2x + 4)]
Refer to Chapter 10 if you need a refresher on binomial factorizations.
29. Completely factor 5x3 – 80x.
Solve It
30. Completely factor y5 – 9y3 + y2 – 9.
Solve It
31. Completely factor 3x5 – 66x3 – 225x.
Solve It
32. Completely factor z6 – 64.
Solve It
33. Completely factor 8a3b2 – 32a3 – b2 + 4.
Solve It
34. Completely factor z8 – 97z4 + 1,296. (Hint: 1,296 = 81 × 16.)
Solve It
35. Completely factor 4m5 – 4m4 – 36m3 + 36m2.
Solve It
36. Completely factor .
Solve It
Answers to Problems on Factoring Trinomials and Other Expressions
This section provides the answers (in bold) to the practice problems in this chapter.
1. Factor out the GCF: . The answer is .
2. Factor out the GCF: 36w4 – 24w3 – 48w2. The answer is 12w2(3w2 – 2w – 4).
3. Factor out the GCF: .
The answer is .
The trinomial in the brackets doesn’t factor as a quadratic-like expression, so you need to expand each term by multiplying and simplifying.
4. Factor out the GCF: .
5. Factor . The answer is , considering 15 and 1, or 5 and 3 for the factors of 15.
6. Factor . The answer is , which needs opposite signs with factors of 40 to be either 40 and 1, 20 and 2, 10 and 4, or 8 and 5.
7. Factor . The answer is .
8. Factor . The answer is (2z + 3)2.
9. Factor . The answer is , which is a difference of squares.
10. Factor . The answer is .
11. Factor . The answer is .
12. Factor . The answer is , which is a difference of squares.
13. Factor . The answer is .
14. Factor . The answer is . Twice is .
15. Completely factor . The answer is .
First, take out the GCF, 3. Then factor the trinomial in the parentheses.
16. Completely factor . The answer is .
The GCF is 5y. Factor that out first.
17. Completely factor . The answer is .
18. Completely factor . The answer is .
This is a quadratic-like expression. After factoring the trinomial, you find that both binomials are the difference of perfect squares.
19. Completely factor . The answer is .
First, divide out the GCF, . Then you have a factorable trinomial in the parentheses.
20. Completely factor .
The GCF is , which is the difference of squares.
21. Completely factor . The answer is .
The quadratic-like trinomial has fractional exponents with the exponent being twice the exponent . The factors of 40 are 1 and 40, 2 and 20, 4 and 10, 5 and 8. The factors of 15 are 1 and 15, 3 and 5. You have to find a combination of factors that gives you a difference of 1 for the middle term. Using the 8 and 5 with the 3 and 5 does the trick.
The product of the outer terms is 24, and the product of the inner terms is 25: (8 5)(5 3).
22. Completely factor . The answer is .
First divide by the GCF to get . You can factor the trinomial as it’s written or rearrange the expression to read . Now the trinomial factors, and you get .
23. Factor by grouping: . The answer is .
24. Factor by grouping: . The answer is .
25. Factor by grouping: . The answer is .
Be sure to factor –4 out of the second two terms to make the binomials match.
Now you factor the difference of squares.
26. Factor by grouping: . The answer is .
The GCF of the first two terms is . The GCF of the last two terms is just 1. So .
27. Factor by grouping: . The answer is .
28. Factor by grouping: . The answer is .
29. Completely factor 5x3 – 80x. The answer is 5x(x + 4)(x – 4). First factor 5x out of each term: 5x(x2 – 16). Then you can factor the binomial as the sum and difference of the same two terms.
30. Completely factor y5 – 9y3 + y3 – 9. The answer is (y + 3)(y – 3)(y + 1)(y2 – y + 1).
31. Completely factor 3x5 – 66x3 – 225x. The answer is 3x(x + 5)(x – 5)(x2 + 3).
32. Completely factor z6 – 64. The answer is (z + 2)(z2 – 2z + 4)(z – 2)(z2 + 2z + 4).
When you do this problem as the difference of squares rather than the difference of cubes, the second part of the factoring is easier.
33. Completely factor 8a3b2 – 32a3 – b2 + 4. The answer is (b + 2)(b – 2)(2a – 1)(4a2 + 2a + 1).
34. Completely factor . The answer is .
35. Completely factor . The answer is .
36. Completely factor . The answer is
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Part III
Seek and Ye Shall Find. . .Solutions
In this part . . .
Everyone likes to be the one to solve the prickly problem that has stumped others, whether it’s to become an instant millionaire or just the hero of the day finding a way around the traffic jam. Finding the solution or solutions of an equation or inequality takes the three P’s: preparation, perseverance, and a plan. You need to come with the proper tools — pencil, paper, and basic algebraic skills — but a solution isn’t going to come to you if you aren’t willing to diligently carry on with a thought-out plan. In these chapters, you see how to recognize the different types of problems and decide on the best approaches. You then can apply the different algebraic techniques and see why they are so important.
Chapter 12
Lining Up Linear Equations