Multiplication and Factoring of Polynomials

Multiplication of Polynomials

  • Overview of Polynomial Multiplication:

    • Multiplying polynomials involves distributing each term of one polynomial to every term of another polynomial and combining like terms.

  • Monomial by Binomial Multiplication:

    • Example Problem: Multiply 8x8x and 6x−46x - 4

    • Step 1: Set up the multiplication:       (8x)(6x−4)(8x)(6x - 4)

    • Step 2: Distribute 8x8x into the binomial:       =48x2−32x= 48x^2 - 32x

  • Additional Monomial Multiplication Examples:

    • Problem (i): Multiply −6r-6r and 2r−52r - 5

    • Step 1: Set up the expression:       (−6r)(2r−5)(-6r)(2r - 5)

    • Step 2: Distribute −6r-6r:       =−12r2+30r= -12r^2 + 30r

    • Problem (ii): Multiply 2p2p and 3p2+53p^2 + 5

    • Step 1: Set up the expression:       (2p)(3p2+5)(2p)(3p^2 + 5)

    • Step 2: Distribute 2p2p:       =6p3+10p= 6p^3 + 10p

    • Problem (iii): Multiply −5y-5y and y3−2y^3 - 2

    • Step 1: Set up the expression:       (−5y)(y3−2)(-5y)(y^3 - 2)

    • Step 2: Distribute −5y-5y:       =−5y4+10y= -5y^4 + 10y

  • Binomial by Binomial Multiplication (The FOIL Method):

    • The FOIL acronym stands for:

    • First: Multiply the first terms in each set of parentheses.

    • Outside: Multiply the outer terms in the expression.

    • Inside: Multiply the inner terms in the expression.

    • Last: Multiply the last terms in each set of parentheses.      

      FOIL Method Breakdown

    • Problem (iv): Multiply (x+2)(x + 2) and (x+5)(x + 5)

    • Step 1: Apply FOIL expansion:

      • First terms: x×x=x2x \times x = x^2

      • Outside terms: x×5=5xx \times 5 = 5x

      • Inside terms: 2×x=2x2 \times x = 2x

      • Last terms: 2×5=102 \times 5 = 10

    • Step 2: Write combined terms:       =x2+5x+2x+10= x^2 + 5x + 2x + 10

    • Step 3: Combine like terms (5x+2x=7x5x + 2x = 7x):       =x2+7x+10= x^2 + 7x + 10

    • Problem (v): Multiply (2x−5)(2x - 5) and (3x+4)(3x + 4)

    • Step 1: Distribute each term of the first binomial across the second binomial:       =2x(3x+4)−5(3x+4)= 2x(3x + 4) - 5(3x + 4)

    • Step 2: Perform individual distributions:       =6x2+8x−15x−20= 6x^2 + 8x - 15x - 20

    • Step 3: Combine like terms (8x−15x=−7x8x - 15x = -7x):       =6x2−7x−20= 6x^2 - 7x - 20

    • Problem (vi): Multiply (2y+3)(2y + 3) and (y−5)(y - 5)

    • Step 1: Distribute each term:       =2y(y−5)+3(y−5)= 2y(y - 5) + 3(y - 5)

    • Step 2: Expand terms:       =2y2−10y+3y−15= 2y^2 - 10y + 3y - 15

    • Step 3: Combine like terms (−10y+3y=−7y-10y + 3y = -7y):       =2y2−7y−15= 2y^2 - 7y - 15

  • Binomial by Polynomial Multiplication:

    • Problem (vii): Multiply (3k−2)(3k - 2) and (k2+5k−4)(k^2 + 5k - 4)

    • Step 1: Distribute each term of (3k−2)(3k - 2) across the trinomial:       =3k(k2+5k−4)−2(k2+5k−4)= 3k(k^2 + 5k - 4) - 2(k^2 + 5k - 4)

    • Step 2: Perform multiplication across all terms:       =3k3+15k2−12k−2k2−10k+8= 3k^3 + 15k^2 - 12k - 2k^2 - 10k + 8

    • Step 3: Group and combine like terms (15k2−2k2=13k215k^2 - 2k^2 = 13k^2 and −12k−10k=−22k-12k - 10k = -22k):       =3k3+13k2−22k+8= 3k^3 + 13k^2 - 22k + 8

    • Problem (viii): Multiply (x+1)(x + 1) and (x2+x+2)(x^2 + x + 2)

    • Step 1: Distribute each term:       =x(x2+x+2)+1(x2+x+2)= x(x^2 + x + 2) + 1(x^2 + x + 2)

    • Step 2: Perform multiplication:       =x3+x2+2x+x2+x+2= x^3 + x^2 + 2x + x^2 + x + 2

    • Step 3: Combine like terms (x2+x2=2x2x^2 + x^2 = 2x^2 and 2x+x=3x2x + x = 3x):       =x3+2x2+3x+2= x^3 + 2x^2 + 3x + 2

    • Problem (ix): Multiply 2p32p^3 and (5p2−3p+3)(5p^2 - 3p + 3)

    • Step 1: Distribute 2p32p^3 into the trinomial:       =2p3(5p2−3p+3)= 2p^3(5p^2 - 3p + 3)

    • Step 2: Perform term-by-term multiplication and exponent addition:       =10p5−6p4+6p3= 10p^5 - 6p^4 + 6p^3

Factoring Overview

  • Definition of Factoring:

    • Factoring is defined as the reverse process of polynomial multiplication.

    • While multiplication transforms factor expressions into an expanded product, factoring breaks an expanded polynomial down into a product of simpler factors.

  • Arithmetic Analogy of Factoring:

    • Multiplication:     6×3=186 \times 3 = 18

    • Factoring 1818 into factor pairs or prime factors:     18=9×218 = 9 \times 2     18=6×318 = 6 \times 3     18=2×3×318 = 2 \times 3 \times 3

Factoring out the Greatest Common Factor

  • Greatest Common Factor (GCF) Concept:

    • Factoring out the greatest common factor involves identifying the highest common numerical factor and highest variable power common to all terms in a polynomial.

  • Demonstration Example:

    • Factor 12p+18q12p + 18q

    • Step 1: Identify the GCF of coefficients 1212 and 1818, which is 66.

    • Step 2: Divide each term by 66 and write as a product:       =6(2p+3q)= 6(2p + 3q)

  • Practice Problems for GCF Factoring:

    • Problem (i): Factor 15m+3015m + 30

    • The GCF is 1515.

    • Expression factored:       =15(m+2)= 15(m + 2)

    • Problem (ii): Factor 20x2+10x−520x^2 + 10x - 5

    • The GCF is 55.

    • Expression factored:       =5(4x2+2x−1)= 5(4x^2 + 2x - 1)

    • Problem (iii): Factor 5y3+15y+20y25y^3 + 15y + 20y^2

    • The GCF is 5y5y.

    • Expression factored:       =5y(y2+3+4y)= 5y(y^2 + 3 + 4y)

    • Rearranged in standard polynomial degree order:       =5y(y2+4y+3)= 5y(y^2 + 4y + 3)

    • Problem (iv): Factor 8x3−9x2+15x8x^3 - 9x^2 + 15x

    • The GCF is xx.

    • Expression factored:       =x(8x2−9x+15)= x(8x^2 - 9x + 15)

    • Problem (v): Factor 15pq+20q2+10q15pq + 20q^2 + 10q

    • The GCF is 5q5q.

    • Expression factored:       =5q(3p+4q+2)= 5q(3p + 4q + 2)

    • Problem (vii): Factor 6m4+9m3−12m26m^4 + 9m^3 - 12m^2

    • The GCF is 3m23m^2.

    • Expression factored:       =3m2(2m2+3m−4)= 3m^2(2m^2 + 3m - 4)

Factoring by Grouping Technique

  • General Formula and Procedure:

    • Grouping is used when a polynomial contains four terms without a single GCF across all terms.

    • General algebraic expression:     ac+ad+bc+bdac + ad + bc + bd

    • Step 1: Group terms into pairs sharing common factors:     (ac+ad)+(bc+bd)(ac + ad) + (bc + bd)

    • Step 2: Factor out the GCF from each binomial pair:     =a(c+d)+b(c+d)= a(c + d) + b(c + d)

    • Step 3: Factor out the common binomial factor (c+d)(c + d):     =(c+d)(a+b)= (c + d)(a + b)

  • Worked Examples of Grouping:

    • Problem 1: Factor ac+ad+bc+bdac + ad + bc + bd

    • Group first two terms and last two terms:       =a(c+d)+b(c+d)= a(c + d) + b(c + d)

    • Factor out binomial (c+d)(c + d):       =(c+d)(a+b)= (c + d)(a + b)

    • Problem 2: Factor 12a3+2a2−30a−512a^3 + 2a^2 - 30a - 5

    • Step 1: Group into pairs (12a3+2a2)+(−30a−5)(12a^3 + 2a^2) + (-30a - 5).

    • Step 2: Factor out 2a22a^2 from the first pair and −5-5 from the second pair:       =2a2(6a+1)−5(6a+1)= 2a^2(6a + 1) - 5(6a + 1)

    • Step 3: Factor out common binomial factor (6a+1)(6a + 1):       =(6a+1)(2a2−5)= (6a + 1)(2a^2 - 5)

    • Problem 3: Factor 4x3−12x2−5x+154x^3 - 12x^2 - 5x + 15

    • Step 1: Group terms (4x3−12x2)+(−5x+15)(4x^3 - 12x^2) + (-5x + 15).

    • Step 2: Factor out 4x24x^2 from first pair and −5-5 from second pair:       =4x2(x−3)−5(x−3)= 4x^2(x - 3) - 5(x - 3)

    • Step 3: Factor out common binomial (x−3)(x - 3):       =(x−3)(4x2−5)= (x - 3)(4x^2 - 5)

    • Problem 4: Factor d2+cd+2c+2dd^2 + cd + 2c + 2d

    • Step 1: Group terms (d2+cd)+(2c+2d)(d^2 + cd) + (2c + 2d):       =d(d+c)+2(c+d)= d(d + c) + 2(c + d)

    • Step 2: Recognize (d+c)(d + c) is equivalent to (c+d)(c + d):       =d(c+d)+2(c+d)= d(c + d) + 2(c + d)

    • Step 3: Factor out common binomial (c+d)(c + d):       =(c+d)(d+2)= (c + d)(d + 2)

    • Problem 5: Factor b3−4b2+b−4b^3 - 4b^2 + b - 4

    • Step 1: Group terms (b3−4b2)+(b−4)(b^3 - 4b^2) + (b - 4):       =b2(b−4)+1(b−4)= b^2(b - 4) + 1(b - 4)

    • Step 2: Factor out common binomial (b−4)(b - 4):       =(b−4)(b2+1)= (b - 4)(b^2 + 1)

    • Problem 6: Factor 5t3+6t2+5t+65t^3 + 6t^2 + 5t + 6

    • Method 1 (Sequential Grouping):

      • Step 1: Group (5t3+6t2)+(5t+6)(5t^3 + 6t^2) + (5t + 6):         =t2(5t+6)+1(5t+6)= t^2(5t + 6) + 1(5t + 6)

      • Step 2: Factor out common binomial (5t+6)(5t + 6):         =(5t+6)(t2+1)= (5t + 6)(t^2 + 1)

    • Method 2 (Alternative Regrouping):

      • Step 1: Rearrange terms to group like coefficients:         =(5t3+5t)+(6t2+6)= (5t^3 + 5t) + (6t^2 + 6)

      • Step 2: Factor out 5t5t from first pair and 66 from second pair:         =5t(t2+1)+6(t2+1)= 5t(t^2 + 1) + 6(t^2 + 1)

      • Step 3: Factor out common binomial (t2+1)(t^2 + 1):         =(t2+1)(5t+6)= (t^2 + 1)(5t + 6)

Factoring Difference of Squares

  • Difference of Squares Formula:

    • Formula rule:     a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)

  • Worked Examples:

    • Problem (1): Factor m2−9m^2 - 9

    • Step 1: Rewrite as difference of squares:       =m2−32= m^2 - 3^2

    • Step 2: Identify terms (a=ma = m, b=3b = 3):       =(m+3)(m−3)= (m + 3)(m - 3)

    • Problem (2): Factor y2−16y^2 - 16

    • Step 1: Rewrite as difference of squares:       =y2−42= y^2 - 4^2

    • Step 2: Identify terms (a=ya = y, b=4b = 4):       =(y+4)(y−4)= (y + 4)(y - 4)

    • Problem (3): Factor 4x2−364x^2 - 36

    • Step 1: Rewrite as difference of perfect squares:       =(2x)2−62= (2x)^2 - 6^2

    • Step 2: Identify terms (a=2xa = 2x, b=6b = 6):       =(2x+6)(2x−6)= (2x + 6)(2x - 6)

    • Problem (4): Factor z2−25z^2 - 25

    • Step 1: Rewrite as difference of squares:       =z2−52= z^2 - 5^2

    • Step 2: Identify terms (a=za = z, b=5b = 5):       =(z+5)(z−5)= (z + 5)(z - 5)

    • Problem (5): Factor x4−y4x^4 - y^4

    • Step 1: Apply difference of squares to fourth powers:       =(x2)2−(y2)2= (x^2)^2 - (y^2)^2       =(x2+y2)(x2−y2)= (x^2 + y^2)(x^2 - y^2)

    • Step 2: Factor the resulting difference of squares (x2−y2)(x^2 - y^2) completely:       =(x2+y2)(x+y)(x−y)= (x^2 + y^2)(x + y)(x - y)

    • Problem (6): Factor 4p4−4q44p^4 - 4q^4

    • Step 1: Factor out GCF of 44:       =4(p4−q4)= 4(p^4 - q^4)

    • Step 2: Apply difference of squares repeatedly:       =4(p2+q2)(p2−q2)= 4(p^2 + q^2)(p^2 - q^2)       =4(p2+q2)(p+q)(p−q)= 4(p^2 + q^2)(p + q)(p - q)

    • Problem (7): Factor 128p2−98q2128p^2 - 98q^2

    • Step 1: Factor out GCF of 22:       =2(64p2−49q2)= 2(64p^2 - 49q^2)

    • Step 2: Rewrite terms as perfect squares:       =2((8p)2−(7q)2)= 2((8p)^2 - (7q)^2)

    • Step 3: Apply difference of squares formula (a=8pa = 8p, b=7qb = 7q):       =2(8p+7q)(8p−7q)= 2(8p + 7q)(8p - 7q)