Beginning Algebra 5.6 Factoring: A General Review

BeginningAlgebra

5.6 Factoring: A General Review

Objective 1. To factor a variety of polynomials.

6.5 Factoring: A General ReviewA General Review

6.5 Factoring: A General ReviewA General Review

1. Check for common factorscommon factors and factor them outfactor them out of the polynomial.

2. If there is no common factorno common factor check for the two special types special types of factorable polynomialsfactorable polynomials:.

a) Difference of SquaresDifference of Squares (binomialbinomial).

b) Perfect Square TrinomialPerfect Square Trinomial

3. Check for factors of general polynomialfactors of general polynomial.

Check for common factorscommon factors andfactor them outfactor them out of the polynomial.

Given a general polynomialgeneral polynomial:

What to do How to do it

25ax3 - 40abx2 + 15a2x2

5ax2(5x - 8b + 3a)

Factor Common Factors: Polynomials

Given a general binomialgeneral binomial: 64a2x2 + 16b2x2

16x2(4a2 + b2)prime factorprime factor

prime factorprime factor



Factor general polynomials:

Given a special binomialspecial binomial 64a2x2 – 16b2x2

16x2(4a2 – b2)difference of squaresdifference of squares


(2a)2 – (b)2 difference of squaresdifference of squares

prime factorsprime factors(2a + b)(2a – b)

Rewrite square factorsquare factor(ss) in the inner binomialinner binomial.

Factor as sumsum and differencedifference of bases of the squaresbases of the squares.

64a2x2 – 16b2x2 =16x2(2a + b)(2a – b)

Multiply by common factorcommon factor 16x16x22

What to Do How to Do It

Given Even Degree BinomialEven Degree Binomial:

(ab)2 – (3c)2 difference of squaresdifference of squares

prime factorsprime factors

Factor Difference of SquaresDifference of Squares:

(ab + 3c)(ab – 3c)

a2b2 – 9c2 =(ab + 3c)(ab – 3c)

Rewrite square factorsquare factor(ss) in each term.


The factorsfactors of differencedifference of squaressquares form the productproduct of conjugate pairsconjugate pairs.

a2b2 – 9c2


3x2 – 12

3(x2 – 4)difference of squaresdifference of squares


Factor binomial as the difference of squaresdifference of squares

Multiply by common factorcommon factor 3 left in composite or power form.


(x + 2)(x – 2)

3x2 – 12 =3(x + 2)(x – 2)

Factor out the common factor(s) from each term.

Given Second Degree BinomialSecond Degree Binomial:


Given Even Degree BinomialEven Degree Binomial: x4 – y8 (x2)2 – (y4)2

difference of squaresdifference of squares



(x2 + y4)(x + y2)(x – y2)

x4 – y8 = (x2 + y4)(x + y2)(x – y2)

Rewrite square factorsquare factor(ss) in each term.

Factor second binomialsecond binomial again.

Completely factor Completely factor the differencedifference of squaressquares requires repeated factoringrepeated factoring.


(x2 + y4)(x2 – y4)difference of squaresdifference of squares

x2 – y4


Given Perfect Square TrinomialPerfect Square Trinomial:

Factor Perfect Square TrinomialPerfect Square Trinomial:

a2x2 2abx + b2 = (ax

b)2

(ax)2 2abx +

b2

ax twicetwice b

2abx

Perfect square trinomials must have the firstfirst and last termslast terms be perfect squares and the last sign positivelast sign positive.

If all of these conditions hold, check to see if the middle term is twicetwice the productproduct of square rootssquare roots of first termfirst term and last last termterm.

The middle signmiddle sign is the sign of the binomialsign of the binomial .

a2x2 2abx + b2

(ax b)2Factors Factors of the square of the binomial with square rootssquare roots of first first and last termslast terms.




25x2 + 60x + 36 = (5x + 6)2

(5x)2 + 60x + 62Perfect square trinomialsPerfect square trinomials must have the firstfirst and last termslast terms be perfect squares and the last sign positivelast sign positive.


The middle signmiddle sign + is the

sign of the binomialsign of the binomial .

25x2 + 60x + 36

(5x + 6)2Factors Factors of the square of the binomial with square rootssquare roots of first first and last termslast terms.

5x twicetwice 6

2 5x 6

60x60x




16x2 – 56x + 49 = (4x – 7)2

(4x)2 – 56x + 72Perfect square trinomialsPerfect square trinomials must have the firstfirst and last termslast terms be perfect squares and the last sign positivelast sign positive.


The middle signmiddle sign – is the


16x2 – 56x + 49

(4x – 7)2Factors Factors of the square of the binomial with square rootssquare roots of first first and last termslast terms.

4x twicetwice 7

2 4x 7

56x56x




81x2 – 90x + 25 = (9x – 5)2

(9x)2 – 90x + 52Perfect square trinomialsPerfect square trinomials must have the firstfirst and last termslast terms be perfect squares and the last sign positivelast sign positive.

If all of these conditions hold, check to see if the middle term is twicetwice the productproduct of square rootssquare roots of first termfirst term and last termlast term.

The middle signmiddle sign – is the


81x2 – 90x + 25

(9x – 5)2Factors Factors of the square of the binomial with square rootssquare roots of first first and last termslast terms.

9x twicetwice 5

2 9x 5

90x90x

Find the productproduct of the firstfirst and last coefficientslast coefficients: AA and CC

Find all of the pairs of factorspairs of factors

rr and ss

Given a general quadratic trinomialgeneral quadratic trinomial:


is the middle coefficient . rr – s = Bs = B or

r + s = Br + s = Bso their sumsum

P = rP = r ss, r > sr > s

Ax2 Bx C

AACC = P

Factor general quadratic trinomial:

or differencedifferenceis the middle coefficient .

Factor by Clue of SignsClue of Signs: What to Do How to Do It

Ax2 Bx C

Read as “+ “+ oror ””

Given general trinomial that has no common factor. See clues of the signsclues of the signs.

P = AP = ACC is the grouping numbergrouping number

(r – s) = Bwhose sumsum or differencedifference is B

++ sumsum

Find all possible factors of GN = PP = rs , r > s

(r + s) = B

– differencedifference

Sum

+

Difference

Same signSame sign

Larger signLarger sign

Rewrite middle termmiddle term BxBx: Ax2 rx rx sx sx + C

underlineunderline and factor by groupingfactor by grouping (ax b)(cx d)

Find all of the all of the pairspairs of factors:

r and s

Find the productproduct of 66 and 1515:

Middle signMiddle sign is –– therefore:

90 · 1

GN: 6 ·15 = 90

–10 , + 9


10 – 9 = 1

18 · 515 · 6

30 · 3

45 · 2

Separate middle termmiddle term –11·tt: –10t , + 9t

with the differencedifference = 1. 10 · 9 90

Example: 6t2 – t – 15

Copy the polynomialpolynomial:

Rewrite middle termmiddle term –– tt: and group for factoringgroup for factoring

Factor each groupeach group:

Factor common factorcommon factor.

6t2 –– 10t + 9t10t + 9t –– 15

6t2 –– t –– 15


(2t + 3)(3t –– 5)

Continue Example: 6t2 – t – 15

2t(3t –– 5) + 3(3t –– 5)

Always Check Factors.Always Check Factors. (2t + 3)(3t –– 5)

See next slide:See next slide:

+bring down middle signmiddle sign

underline common factor:

Check Factors using FOIL

6t2 – t – 15= (2t + 3)(3t – 5)


Check:Check: Multiply using

First

Note sum of O + I terms

F 0 I L

(2t + 3)(3t - 5)

Outer

Inner

Last6t2 –– 10t + 9t –– 15

6t2 - 15 - 10t

+ 9t

Always Check Factors.Always Check Factors.

6t2 –– 1t –– 15


Ax2 Bx C

k·(ax2 bx

c)

k·ax2 k·bx

k·c

FactorFactor out the common factor(ss) from each termeach term.

Apply the distributive propertydistributive property.

Trinomials with Common FactorsCommon Factors:

As common factorscommon factors numbersnumbers are left in composite formcomposite form andand lettersletters are left in power formpower form.

Check Inner Polynomial for Clue of SignsClue of Signs and GN

axax22 bx bx

cc


72x2 –– 60x –– 28

4(18x2 – 15x – 7)

4·18x2 – 4·15x –

4·7

Factor out the common factor(ss) from each term.

Apply the distributive propertydistributive property.

Trinomials with Common FactorsCommon Factors:

Numbers as common factorscommon factors are left in composite formcomposite form.

Check: Inner Polynomial for Clue of SignsClue of Signs and GN

18x2 – 15x – 7

Hold the 4Hold the 4

Inner TrinomialInner Trinomial: 18x18x22 - 15x - 7 - 15x - 7

Find all of the pairs of factors: r and s

Find the productproduct of 1818 and 77:

Middle sign is –– therefore:

18 · 7 = 126

- 21 , + 6


Separate middle termmiddle term - 15x- 15x: - 21x , + 6x

with the difference difference = 15.

126 · 1

21 - 6 = 15

21 · 618 · 7

42 · 3

63 · 2

14 · 9126

Copy the polynomialpolynomial:

Rewrite middle termmiddle term -15x -15x : and group for factoringgroup for factoring

Factor each groupeach group:

Factor common factorcommon factor.

18x2 – 21x– 21x + 6x+ 6x –– 7

18x2 –– 15x –– 7


(3x + 1)(6x –– 7)

Continue Example: 18x2 –– 15x –– 7

3x(6x - 7) + 1(6x - 7)

Always Check Factors.Always Check Factors. (3x + 1)(6x –– 7)

See next slide:See next slide:

+bring down middle signmiddle sign

underline common factor:

Check Factors by FOIL What to Do How to Do It

Check factors of inner trinomialinner trinomial by

First

Find the sum of O + I terms

F 0 I L

(3x + 1)(6x –– 7)

18x2 –– 15x –– 7

Outer

Inner

Last 18x2 –– 21x + 6x –– 7

18x2 - - 7-- 21x

+ 6x

Multiply by common factorcommon factor 4472x2 –– 60x –– 28 =

4(3x + 1)(6x –– 7)

Given a quadratic trinomialquadratic trinomial and integers integers a, b, c

The test requires knowing squaressquares of integers or use to a calculator.


Review squares Review squares of integersof integers

ax2 + bx + c

Quadratic Trinomial Test for FactorabilityTest for Factorability:

The trinomial will factor with have rational factorsrational factors if:

b2 –– 4ac = d2

a > 0a > 0signs included in signs included in b b and and cc

dd 0 0

b2 –– 4ac =0 perfect square trinomialperfect square trinomial.

d2 factorable trinomialfactorable trinomial.d2 = +d


Test 4x2 –– 12x - 9 4x2 –– 12x –– 9

(-12)2 –– 4(4)(-9) = 288not factorablenot factorable

a = 4, b = -12, c = -9a = 4, b = -12, c = -9

4x2 –– 12x - 9 =

(2a –– 3)2

bb22 –– 4ac = 04ac = 0

d2

Test 4x2 –– 12x + 9 4x2 –– 12x + 9

(-12)2 –– 4(4)(9)

perfect square trinomialperfect square trinomial

a = 4, b = -12, c = 9a = 4, b = -12, c = 9 = 0

Perfect square trinomial Perfect square trinomial

yields square of binomialsquare of binomial



Test 4x2 +15x + 9 4x2 + 15x + 9(-15)2 –– 4(4)(9) = 81

factorable trinomialfactorable trinomiala = 4, b = -15, c = 9a = 4, b = -15, c = 9

4x2 + 15x + 9 =

(4x + 3)(x + 3)

Sum:Sum: 12 + 3 = 15 12 + 3 = 15

= 92

FactorFactor 4x2 + 15x + 9 4x2 + 15x + 9

4x(x + 3) + 3(x + 3)

GN = 36GN = 36 -- -- factorsfactors 12 and 3 12 and 3

CompleteComplete the factors:


4x2 + 12x + 3x + 9

Perfect square trinomialPerfect square trinomial.

Factor: common factorscommon factors:


ax + ay = a(x + y)

a2x2 2abx + b2

Summary: Factor PolynomialsFactor Polynomials

Difference of SquaresDifference of Squares: A2 –– B2 = (A + B)(A –– B)

(ax b)2

Distribute: left / rightleft / right ax + bx = (a + b)x

Conjugate PairsConjugate Pairs: a2x2 –– b2y2 =

binomial squaredbinomial squared

(ax+by)(ax –– by)

Difference Difference ofof like evenlike even powerspowersWhat to do How to do it

xn - yn = (x2m –– y2m)

(xm + ym)(xm –– ym)


Repeat since m = 4, p = 2m = 4, p = 2 (xm + ym) is prime

(x4+y4)(x2+ y2)(x2 –– y2)

(x4 + y4)(x4 –– y4)

[(x2)2 –– (y2)2]n = 4, m = 2:

Repeat since m = 2m = 2 (xm + ym) is prime

(x2 + y2)(x2 –– y2)

(x2 + y2)(x + y)( x –– y)

n = 8, m = 4, p = 2:

x4 – y4 =

x8 – y8 =

Repeat if m = 2pm = 2p (xm + ym) is prime

(x4+y4)(x2+ y2)(x+ y)(x –– y)

Factor by groupingFactor by grouping.


Apply commutative property


ax + by - bx - ay

= (a (a – b)(x b)(x – y) y)

PolynomialsPolynomials of of 44 or more or more terms rearranged / groupedterms rearranged / grouped:

= ax – ay – bx + by

= a(x – y) – b(x – y)Apply distributive property:

Underline common factor:

Apply distributive property

= a(x – y) – b(x – y)

Underline groupsUnderline groups:

THEEND

5.6

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Beginning Algebra 5.6 Factoring: A General Review