5 11 Using Several Methods of Factoring

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5—11 Using Several Methods of Factoring

Objective: To factor polynomials completely.

Vocabulary

Factored completely A polynomial expressed as the product of a monomial and one or more prime polynomials, that is when it cannot be factored further.

Guidelines for Factoring Completely

1. Factor ut the greatest monomial factor first. 2. Look for a difference of squares.

Pattern: a2 — b2 = (a — b)(a + b) (However, a2 + b2 can’t be factored.) 3. Look for a perfect square thnomial.

Patterns: (a+b)2=a2+2ab+b2 (a—b)2=a2—2ab+b2

4. If the trinomial is not a perfect square, look for a pair of binomial factors. 5. If a polynomial has four or more terms, look for a way to group the terms

in pairs or in a group of three terms that is a perfect square tñnomial. 6. Make sure that each binomial or tflnomial factor is prime.

7. Check your work by multiplying the factors.

Example 1 Factor 8x3 — 512x completely.

Solution 8x3 — 512x = Sx(x2 — 64)

Greatest monomial factor

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Difference of squares =Sx(x+8)(x—8) Factor completely 1 3x3—12x 2 5m3—45m 3 3a2+6ab+3b2 I —x3+4xy2 S —12z3+30z2+lSz 6 16r4—24r3+9r2 S20x3_28x2+8x 8 j3+j2—2t 9.2x2—128 1O.2x4—162 H. 25z3—36yz . 12. &+22ay—8y2

StudyGuide, ALGEBRA, Structure and Method, Book 1

-Example 2

Solution

Factor 3x3 + 3x2 — 18x completely. 3x3 + 3? — 18x = 3x(? +x — 6)

Greatest monomial factor Trinomial

3x(x + 3)(x — 2) S.

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5—11 Using Several Methods of Factoring (continued)

DATE

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Example 3 Factor 5a2b3 2a3b2 — 3ab4 completely.

Solution First rewrite the polynomial in order of decreasing degree in a:

. 5a2b3 + 2a3b2 — 3ab4 = 2a3b2 + 5a2b3 — 3ab4

. ₌ ab2(2a2+Sab_—_3b2)

Greatest monomial factor • Trinomial

. =ab2(2a—b)(a+3b)

ExampIe 4 Factor a2b — 4b + 3a2 — 12 completely.

Solution a2b — 4b + 3a2 — 12 = b(a2 — 4) ± 3(a2 — 4) Group and factor.

= (b + 3)(a2 — 4) Use the distributive property. t _{Difference of squares} I =(b+3)(a+2)(a—2) Factor completely. 13.a3x—9ax3 . Q18x3_24x2+8x 15. 20 — 60x + 45.? 16. 6x2 — l8xy + 12y2 17. 9x3+108x+63x2 . . .10k3+25k_.35k2

19..32A—48r3+ 18? 20. 12ab—3b2— 12a2

G&2_4b+3c2_12 22.x3—x+6x2—6

23.x2+6xy+9y2—16 . 24.a3+a2b—ab2—b3

25.y4—9y2+20 2&x4—ltk2+9

27.x4—13x2+36 28.x’t—’24x2+144 .

29.b4—8b2+16 _{@a3+2a2_5a_10}

Mixed Review Exercises

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5—12 Solving Equations by Factoring

Objective: To use factoring in solving polynomial equations.

Vocabulary

property A product of factors is zero if and only if one or more of the factors is zero.

Polynomial equation An &quation whose sides are both polynomials.

Linear equation .A polynomial equation whose term of highest degree has degree 1. For example, x — 2 = 0 and 5x — 4 = 6.

Quadratic equation A polynomial equation whose term of highest degree has degree 2. For example, x2 — x — 6 = 0, x2 = 9x, and lOx— 9 = x2.

Cubic equation A polynomial dquation whose term of highest degree has degree 3. For example, x3 — 2x2 + x — 1 = 0.

Standard form of a polynomial equation A form of an equation in which one side is a simplified polynomial arranged in order of decreasing degree of the variable and the other side is zero.

Double or multiple root

A

factor that occurs twice in the factored form of an equation. For example, 5 is a doubie root of x(x — 5)(x — 5) = 0.

Example 1 Solve (x — l)(x + 3) = 0.

Solution Since the product of factors is 0, one of the factors on the left side must equal 0.

. x—l=0 or x+3=0

- x=1 x=—3

The solution set is

_{

1,— 3}. Just by looking at the original equation, you can see that when x = 1 or x = —3, the product will be 0.

Example 2 Solve 3n(n — 2)(n — 5) = 0.

Solution 3,i=0 or n—2=0or .n—5=0

rz=0 n=2 n=5 Thesolutionsetis{0,2,5}.

CAUTION _{Never transform an equation by dividing by an expression containing a}

variable. Notice that in Example 2, the solution 0 would have been lost if both sides of the equation had been divided by 3n.

Solve. l.(y+4)Q’—5)=O. 2.0=(n+l)(n+8) 3. lOn(n—2)=0 4. 2xx — 10) = 0 5. (p — 1)(p — 7) = 0 6. 0 2n(n —

00’

— 3) 7. x(2x — 1)(2x + 1) = 0 8. 0 = n(n — 6) 9. 0 = 3x(4x - l)(x — 2) NAME DATE ____ _____ L

Study Guide, ALGEBRA, Structure and Method, Book I

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NAME

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5—12 Solving Equations by Factoring

(continued)

Example 3 Solve the quadratic equation 2x2 — x = 3. .

Solution 1. Transform the equation into standard form. 2x2 — x — 3 = 0

2. Factor the left side. (2x — 3)(x + 1) = 0

-. 3. Seteachfactorequaltooandsolve. 2x —3 = 0 or x + 1 =0

2x=3 x=—1

3.

4.

Cheek the solutions in the original equation. .

2(i-)2_E23 2(_02_(_1)23 9

2N123

2(1)+13 1\4J 2 3=3 9_.i3 2 2 9 3_6_Q _I

The solution set is {—i

f}

Solve. . 10. x2 — x — 12 = 0 Qx2 — 12x + 27= 0 12. 0 = — 4x — 32 13. 0 = m2 + 3m — 54 14. x2 — 4y + 3 = 0 15. x2 — lOx — 24 = 0 16..0=n2—n 17.y2=12y 18.6k2=2k 19. x2 + 16 = 8x 20. a2 = 10 — 30 21. 3? — x = 2

6o

= x2 + 12x + 35 23. y2 + 5y = 14 24. x2 = 5x + 36 25.4m2—25=O .. 26.r2+8=9r 27.6n2—n=2 28.3x2+1=4x 29.3a2=6a _{30.3p2—14p=80} 2x2=10+x 32.3p2+17p=-1O 93x2+l=4x

Mixed Review Exerciseé

Evaluate if x = 3 and y = 6.

1. (x—y)3 2.x3’x2 3. 4x3

4.(4x)3 5.3x+y2 6.3x2+y

7.3(x+y)2 8.(yx)2 . 9y2x2

Simplify. . . ..

10. (5x2y2)(—3xy4) 11. (8a) 12. —3(x + 4) .

....

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Study Guide, ALGEBRA, Structure and MethodBook 1

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5—13 Using Factoring to Solve Problems

Objective: To solve problems by writing and factoring quadratic equations.

CAUTION •A solution of an equation may not satisf5i some of the conditions of the problem. You reject solutions of an equation that do not make sense for the problem.

Example I

Solution

Find two consecutive positive odd integers whose product is 143.

Step] Step 2 Step 3 Step 4

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The problem asks for two consecutive positive odd integers.

Let n the first integer. Then ii + 2 = the second integer.

Use the facts in the problem to write an equation. n(n + 2) = 143 Solve the equation. + 2n — 143

(ii + 13)(n — 11) n — 11 ‘I n + 13 = 0 or n = —13 =0 =0 =0 = 11

You are to find positive odd integers, so reject —13. If n = StepS Check: 11 x 13 = 143. The integers are 11 and 13.

Example 2

11, thenn + 2 = 13.

Solution

Originally

a

rectangle was 8 cm jy 17 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 66 cm2. Find the dimensions of the new rectangle.

Step 1 Step 2

The problem asks for the dimensions of the new rectangle.

Let x = the amount by which each dimension is decreased. Make a sketch. The new dimensions are 17 — x and 8 — x.

SIep 3 (Original“ — (Decrease’ = (New area I \ inarea I area (17 8) Step 4 66 = (17—x)(8—x) 136 — 66 70 0 -0 0 136 — 25x + = 136—25x+? = 66 — 25x + = — 25x + 66 = (x — 3)(x — 22) l7—x x 8 —x x—3=0 x3 or x—220 = 22 8

Step 5 Check in the words of the problem and The new rectangle is 14 cm long and 5

17

you’ll see that you must reject 22., cm wide.

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5—13

Using Factoring to Solve Problems

(continued) Solve.

1. If a number is added to its square, thetesult is 72. Find the number. 2. If a number is subtracted from its square, the result is 90. Find the numbeé.

JA positive number is 56 less than its square. Find the number.

4. A negative number is 56 less than its square. Find the number.

/S)Find two consecutive negative integers whose product is 72.

6. Find two consecutive positive even integers whose product is 120. sum of the squares of two consecutive positive odd integers is 202. Find the integers.

8. The sum of the squares of two consecutive negative odd integers is 130. Find the integers.

9. The length of a rectangle is 5 cm greater than its width. Find the - dimensions of the rectangle if its area is 126 cm2.

The length of a rectangle is 8 cm less than twice its width. Find the dimensions of the rectangle if the area is 120 cm2.

11. Find the dimensions of a rectangle whose perimeter is 40 m and whose area is 96 m2. (Hint: Let the wi4th be w. Use the perimeter to find the length in terms of w.)

12. Find the dimensions of a rectangle whose perimeter is 52 m and whose area is 160 m2.

13. Originally the dimensions of a rectangle were 12 cm by 7 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 34 cm2. Find the dimensions of the new rectangk.

/i)Originally a rectangle was twice as long as it was wide. When 5 m

was subtracted from its length and 3 m subtracted from its width, the resulting rectangle had an area of 55 m2. Find the dimensions of the

new rectangle. - -.

Mixed Review Exercises

Simplify.

1. (6ab2)(3a2b) - 2. (4a2)3 3. 2a(3 — 2a)

4. (8r(4r?) 5. (2by2) _{6. (-)u8n} —

3Op)

7. (2a + 3)(2a2 + 3 — 5ä) 8. (—20x — 15y)(—%) 9. (5m

— 3)2 Factor completely. 10. ó,? + 2m — 20 11. 28a3 — 7ab2 13. y4 — y3 — 6y2 14. 12rn2 + l3mn + 3n2 r

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12. 15n2+14n—8 1S.8?—14x+3