INTRODUCTION TO FACTORING POLYNOMIALS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice Levasseur

Definitions Recall: Factors of a number are the numbers that divide the original number evenly. Writing a number as a product of factors is called a factorization of the number. The prime factorization of a number is the factorization of that number written as a product of prime numbers. Common factors are factors that two or more numbers have in common. The Greatest Common Factor (GCF) is the largest common factor.

Ex: Find the GCF(24, 40). Prime factor each number:  24 = 2*2*2*3 = 2 3 *  40 = 2*2*2*5 = 2 3 *5  GCF(24,40)= 2 3 = 8

The Greatest Common Factor of terms of a polynomial is the largest factor that the original terms share Ex: What is the GCF(7x 2, 3x) 7x 2 = 7 * x * x 3x = 3 * x The terms share a factor of x  GCF(7x 2, 3x) = x

Ex: Find the GCF(6a 5,3a 3,2a 2 ) 6a 5 = 2*3*a*a*a*a*a 3a 3 = 3*a*a*a 2a 2 = 2*a*a The terms share two factors of a  GCF(6a 5,3a 3,2a 2 )= a 2 Note: The exponent of the variable in the GCF is the smallest exponent of that variable the terms

Definitions To factor an expression means to write an equivalent expression that is a product To factor a polynomial means to write the polynomial as a product of other polynomials A factor that cannot be factored further is said to be a prime factor (prime polynomial) A polynomial is factored completely if it is written as a product of prime polynomials

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the polynomial a sum/difference of cubes? a 3 + b 3 or a 3 – b 3 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

Ex: Factor 7x 2 + 3x Think of the Distributive Law: a(b+c) = ab + ac  reverse it ab + ac = a(b + c) Do the terms share a common factor? What is the GCF(7x 2, 3x)? Recall: GCF(7x 2, 3x) = x  7x 2 +3x=x( + )What’s left? xx Factor out  7x 2 + 3x = x(7x + 3)

Ex: Factor 6a 5 – 3a 3 – 2a 2 Recall: GCF(6a 5,3a 3,2a 2 )= a 2 6a 5 – 3a 3 – 2a 2 = a 2 ( - - ) a2a2 a2a2 a2a2 31 6a 3 3a2  6a 5 – 3a 3 – 2a 2 = a 2 (6a 3 – 3a – 2)

Your Turn to Try a Few

Ex: Factor x(a + b) – 2(a + b) Always ask first if there is common factor the terms share... x(a + b) – 2(a + b) Each term has factor (a + b)  x(a + b) – 2(a + b)= (a + b)( – ) (a + b) x2  x(a + b) – 2(a + b) = (a + b)(x – 2)

Ex: Factor a(x – 2) + 2(2 – x) As with the previous example, is there a common factor among the terms? Well, kind of... x – 2 is close to 2 - x... Hum... Recall: (-1)(x – 2) =- x + 2 = 2 – x  a(x – 2) + 2(2 – x) =a(x – 2) + 2((-1)(x – 2)) = a(x – 2) + (– 2)(x – 2) = a(x – 2) – 2(x – 2)  a(x – 2) – 2(x – 2) =(x – 2)( – ) (x – 2) a2

Ex: Factor b(a – 7) – 3(7 – a) Common factor among the terms? Well, kind of... a – 7 is close to 7 - a Recall: (-1)(a – 7) =- a + 7 = 7 – a  b(a – 7) – 3(7 – a) =b(a – 7) – 3((-1)(a – 7)) = b(a – 7) + 3(a – 7)  b(a – 7) + 3(a – 7) =(a – 7)( + ) (a – 7) b3

Your Turn to Try a Few

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the polynomial a sum/difference of cubes? a 3 + b 3 or a 3 – b 3 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

Factor by Grouping If the polynomial has four terms, consider factor by grouping 1.Factor out the GCF from the first two terms 2.Factor out the GCF from the second two terms (take the negative sign if minus separates the first and second groups) 3.If factor by grouping is the correct approach, there should be a common factor among the groups 4.Factor out that GCF 5.Check by multiplying using FOIL

Ex: Factor 6a 3 + 3a 2 +4a + 2 Notice 4 terms... think two groups: 1 st two and 2 nd two Common factor among the 1 st two terms?  6a 3 + 3a 2 = 3a 2 ( + ) GCF(6a 3, 3a 2 )= 3a 2 3a 2 2a 1 1 Common factor among the 2 nd two terms? GCF(4a, 2)= 2  4a + 2 = 2( + ) a1 Now put it all together...

6a 3 + 3a 2 +4a + 2 =3a 2 (2a + 1) + 2(2a + 1) Four terms  two terms. Is there a common factor? Each term has factor (2a + 1) 3a 2 (2a + 1) + 2(2a + 1) = (2a + 1)( + ) (2a + 1) 3a 2 2 6a 3 + 3a 2 +4a + 2 =(2a + 1)(3a 2 + 2)

Ex: Factor 4x 2 + 3xy – 12y – 16x Notice 4 terms... think two groups: 1 st two and 2 nd two Common factor among the 1 st two terms?  4x 2 + 3xy = x( + ) GCF(4x 2, 3xy)= x xx 4x 3y 4x3y Common factor among the 2 nd two terms? GCF(-12y, - 16x)= -4  -12y – 16x = - 4( ) -4 3y 4x 3y+ 4x Now put it all together...

4x 2 + 3xy – 12y – 16x =x(4x + 3y) – 4(4x + 3y) Four terms  two terms. Is there a common factor? Each term has factor (4x + 3y) x(4x + 3y) – 4(4x + 3y) = (4x + 3y)( ) (4x + 3y) x – 4 4x 2 + 3xy – 12y – 16x =(4x + 3y)(x – 4)

Ex: Factor 2ra + a 2 – 2r – a Notice 4 terms... think two groups: 1 st two and 2 nd two Common factor among the 1 st two terms?  2ra + a 2 = a( + ) GCF(2ra, a 2 )= a aa 2ra Common factor among the 2 nd two terms? GCF(-2r, - a)= -1  -2r – a = - 1( ) 2r+ a Now put it all together...

2ra + a 2 –2r – a =a(2r + a) – 1(2r + a) Four terms  two terms. Is there a common factor? Each term has factor (2r + a) a(2r + a) – 1(2r + a) = (2r + a)( ) (2r + a) a – 1 2ra + a 2 – 2r – a =(2r + a)(a – 1)

Your Turn to Try a Few

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

Special Polynomials  Is the polynomial a difference of squares? a 2 – b 2 = (a – b)(a + b)  Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2

Ex: Factor x 2 – 4 Notice the terms are both perfect squares x 2 = (x) 2 4 = (2) 2  x 2 – 4 = (x) 2 – (2) 2 a 2 – b 2 and we have a difference = (x – 2)(x + 2)  difference of squares = (a – b)(a + b) factors as

Ex: Factor 9p 2 – 16 Notice the terms are both perfect squares 9p 2 = (3p) 2 16 = (4) 2  9a 2 – 16 = (3p) 2 – (4) 2 a 2 – b 2 and we have a difference = (3p – 4)(3p + 4)  difference of squares = (a – b)(a + b) factors as

Ex: Factor y 6 – 25 Notice the terms are both perfect squares y 6 = (y 3 ) 2 25 = (5) 2  y 6 – 25 = (y 3 ) 2 – (5) 2 a 2 – b 2 and we have a difference = (y 3 – 5)(y 3 + 5)  difference of squares = (a – b)(a + b) factors as

Ex: Factor 81 – x 2 y 2 Notice the terms are both perfect squares 81 = (9) 2 x 2 y 2 = (xy) 2  81 – x 2 y 2 = (9) 2 – (xy) 2 a 2 – b 2 and we have a difference = (9 – xy)(9 + xy)  difference of squares = (a – b)(a + b) factors as

Your Turn to Try a Few

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the polynomial a sum/difference of cubes? a 3 + b 3 or a 3 – b 3 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

FOIL Method of Factoring Recall FOIL –(3x + 4)(4x + 5) = 12x x + 16x + 20 = 12x x + 20 The product of the two binomials is a trinomial The constant term is the product of the L terms The coefficient of x, b, is the sum of the O & I products The coefficient of x 2, a, is the product of the F terms

FOIL Method of Factoring 1.Factor out the GCF, if any 2.For the remaining trinomial, find the F terms (__ x + )(__ x + ) = ax 2 3.Find the L terms ( x + __ )( x + __ ) = c 4.Look for the outer and inner products to sum to bx 5.Check the factorization by using FOIL to multiply

Ex: Factor b 2 + 6b there is no GCF 2. the lead coefficient is 1  (1b )(1b ) 3. Look for factors of 51, 5 & 5, 1 (b + 1)(b + 5) or (b + 5)(b + 1) 4. outer-inner product? (b + 1)(b + 5)  5b + b = 6b or (b + 5)(b + 1)  b + 5b = 6b Either one works  b 2 + 6b + 5 = (b + 1)(b + 5) 5. check: (b + 1)(b + 5) =b 2 + 5b + b + 5 = b 2 + 6b + 5

Ex: Factor y 2 + 6y – there is no GCF 2. the lead coefficient is 1  (1y )(1y ) 3. Look for factors of – 55 1, -55 & 5, - 11 & 11, - 5 & 55, - 1 (y + 1)(y – 55) or (y + 5)(y - 11) or ( y + 11)(y – 5) or (y + 55)(y – 1) 4. outer-inner product? (y + 1)(y - 55)  -55y + y = - 54y (y + 55)(y - 1)  -y + 55y = 54y  y 2 + 6y - 55 = (y + 11)(y – 5) 5. check: (y + 11)(y – 5) =y 2 – 5y + 11y - 55 = y 2 + 6y – 55 (y + 5)(y - 11)  -11y + 5y = -6y (y + 11)(y - 5)  -5y + 11y = 6y

Factor completely – 3 Terms Always look for a common factor –immediately take it out to the front of the expression all common factors –show what’s left inside ONE set of parenthesis Identify the number of terms. If there are three terms, and the leading coefficient is positive: –find all the factors of the first term, find all the factors of the last term –Within 2 sets of parentheses, place the factors from the first term in the front of the parentheses place the factors from the last term in the back of the parentheses –NEVER put common factors together in one parenthesis. –check the last sign, if the sign is plus: use the SAME signs, the sign of the 2nd term if the sign is minus: use different signs, one plus and one minus –“smile” to make sure you get the middle term multiply the inner most terms together then multiply the outer most terms together, and add the two products together.

Factor completely: 2x 2 – 5x – 7 Factors of the first term: 1x & 2x Factors of the last term: -1 & 7 or 1 & -7 (2x – 7)(x + 1)

Factor Completely. 4x x + 60 Nothing common Factors of the first term: 1 & 4 or 2 & 2 Factors of the last term: 1,6 2,30 3,20 4,15 5,12 6,10 Since each pair of factors of the last has an even number, we can not use 2 & 2 from the first term (4x + 3)(1x + 20 )

Sign Pattern for the Binomials Trinomial Sign PatternBinomial Sign Pattern + + ( + )( + ) - + ( - )( - ) plus and 1 minus plus and 1 minus But as you can tell from the previous example, the FOIL method of factoring requires a lot of trial and error (and hence luck!)... Better way?

Your Turn to Try a Few

ac Method for factoring ax 2 + bx + c 1.Factor out the GCF, if any 2.For the remaining trinomial, multiply ac 3.Look for factors of ac that sum to b 4.Rewrite the bx term as a sum using the factors found in step 3 5.Factor by grouping 6.Check by multiplying using FOIL

Ex: Factor 3x 2 – 4x – Is there a GCF? No 2. Multiply ac  a =3 3 and c =– 15  3(-15) = Factors of -45 that sum to – 4 4. Rewrite the middle term 3x 2 – 4x – 15 = 3x 2 – 9x + 5x – 15 1 – 45  – 44 3 – 15  – 12 5 – 9  – 4 Note: although there are more factors of – 45, we don’t have to check them since we found what we were looking for! Four-term polynomial... How should we proceed to factor?

Factor by grouping... 3x 2 – 9x + 5x – 15 Common factor among the 1 st two terms?  3 x 2 – 9x = 3x( – ) 3x 3 x 3 Common factor among the 2 nd two terms? 5  5 x – 15 = 5( – ) 55 3 x3  3x 2 – 9x + 5x – 15 = 3x(x – 3) + 5(x – 3) = (x – 3)( ) 3x + 5

Ex: Factor 2t 2 + 5t – Is there a GCF? No 2. Multiply ac  a =2 2 and c =– 12  2(-12) = Factors of -24 that sum to Rewrite the middle term 2t 2 + 5t – 12 = 2t 2 – 3t + 8t – 12 1 – 24  – 23 2 – 12  – 10 3 – 8  – 5 Four-term polynomial... Factor by grouping... Close but wrong sign so reverse it  5

2t 2 – 3t + 8t – 12 Common factor among the 1 st two terms?  2 t 2 – 3t = t( – ) t t t 3 2t 3 Common factor among the 2 nd two terms? 4  8 t – 12 = 4( – ) t3  2t 2 – 3t + 8t – 12 = t(2t – 3) + 4(2t – 3) = (2t – 3)( ) t + 4 2

Ex: Factor 9x x Is there a GCF? No 2. Multiply ac  a =9 9 and c =8 8  9(8) = Factors of 72 that sum to Rewrite the middle term 9x x = 9x 4 + 6x x    18 Four-term polynomial... Factor by grouping... Bit big  think bigger factors

9x 4 + 6x x Common factor among the 1 st two terms?  9x 4 + 6x 2 = 3x 2 ( + ) 3x Common factor among the 2 nd two terms? 4  12x = 4( + ) x 2 2  9x 4 + 6x x = 3x 2 (3x 2 + 2) + 4(3x 2 + 2) = (3x 2 + 2)( + ) 3x 2 4 3

Ex: Factor 12x 2 – 17xy + 6y 2 1. Is there a GCF? No, but notice two variables 2. Multiply ac  a =12x 2 12 and c =6y 2 12x 2 (6y 2 ) = 72y 2 3. Factors of 72x 2 y 2 that sum to – 17 y - 17xy 4. Rewrite the middle term 12x 2 – 17xy + 6y 2 = 12x 2 – 8xy – 9xy + 6y 2 -1xy -72xy  -73xy -6xy -12xy  -18xy -8xy -9xy  -17xy Four-term polynomial... Factor by grouping... Each factor need a y, both need to be negative Too big, think bigger factors Pick one to be the variable

12x 2 – 8xy – 9xy + 6y 2 Common factor among the 1 st two terms?  12x 2 – 8xy = 4x( – ) 4x 2y 3x 2y Common factor among the 2 nd two terms? - 3y  – 9xy + 6y 2 = - 3y( ) -3y -2y 3x – 2y  12x 2 – 8xy – 9xy + 6y 2 = 4x(3x – 2y) – 3y(3x – 2y) = (3x – 2)( ) 4x– 3y 3 3x

Your Turn to Try a Few

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the polynomial a sum/difference of cubes? a 3 + b 3 or a 3 – b 3 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

Ex: Factor x 3 + 3x 2 – 4x – Is there a GCF? No 2. Notice four terms  grouping Common factor among the 1 st two terms? x2 x2  x 3 + 3x 2 = x 2 ( + ) x2 x2 x2 x2 x x 3 Common factor among the 2 nd two terms? - 4  – 4x – 12 = – 4( ) x + 3  x 3 + 3x 2 - 4x – 12 = x 2 (x + 3) – 4(x + 3) = (x + 3)( ) x 2 – 4

Cont: we have (x + 3)(x 2 – 4) But are we done?No. We have to make sure we factor completely. Is (x + 3) prime?  can x + 3 be factored further? No... It is prime What about (x 2 – 4)? Recognize it? Difference of Squares x 2 = (x) 2 4 = (2) 2  x 2 – 4 = (x) 2 – (2) 2 = (x – 2)(x + 2) Therefore x 3 + 3x 2 – 4x – 12 = (x + 3)(x 2 – 4) = (x + 3)(x – 2)(x + 2)

Your Turn to Try a Few

To factor a polynomial completely, ask Do the terms have a common factor (GCF)? Does the polynomial have four terms? Is the polynomial a special one? –Is the polynomial a difference of squares? a 2 – b 2 –Is the polynomial a sum/difference of cubes? a 3 + b 3 or a 3 – b 3 –Is the trinomial a perfect-square trinomial? a 2 + 2ab + b 2 or a 2 – 2ab + b 2 Is the trinomial a product of two binomials? Factored completely?

Special Polynomials  Is the polynomial a sum/difference of cubes? a 3 + b 3 = (a + b)(a 2 - ab + b 2 ) a 3 – b 3 = (a - b)(a 2 + ab + b 2 )

Ex: Factor 8p 3 – q 3 Notice the terms are both perfect cubes 8p 3 = (2p) 3 q 3 = (q) 3  8p 3 – q 3 = (2p) 3 – (q) 3 a 3 – b 3 and we have a difference = (2p – q)((2p) 2 + (2p)(q) + (q) 2 )  difference of cubes = (a – b)(a 2 + ab + b 2 ) factors as = (2p – q)(4p 2 + 2pq + q 2 )

Ex: Factor x y 9 Notice the terms are both perfect cubes x 3 = (x) 3 27y 9 = (3y 3 ) 3  x y 9 = (x) 3 + (3y 3 ) 3 a 3 + b 3 and we have a sum = (x + 3y 3 )((x) 2 - (x)(3y 3 ) + (3y 3 ) 2 )  sum of cubes = (a + b)(a 2 - ab + b 2 ) factors as = (x + 3y 3 )(x 2 – 3xy 3 + 9y 6 )