Factoring Polynomials Digital Lesson Factoring Polynomials

Greatest Common Factor The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term. Example: Factor 18x3 + 60x. 18x3 = 2 · 3 · 3 · x · x · x Factor each term. = (2 · 3 · x) · 3 · x · x 60x = 2 · 2 · 3 · 5 · x = (2 · 3 · x) · 2 · 5 GCF = 6x Find the GCF. 18x3 + 60x = 6x (3x2) + 6x (10) Apply the distributive law to factor the polynomial. = 6x (3x2 + 10) Check the answer by multiplication. 6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Greatest Common Factor

A common binomial factor can be factored out of certain expressions. Example: Factor 4x2 – 12x + 20. 4x2 = 2 · 2 · x · x , 12x = 2 · 2 · 3 · x , 20 = 2 · 2 · 5 Therefore, GCF = 4. 4x2 – 12x + 20 = 4x2 – 4 · 3x + 4 · 5 = 4(x2 – 3x + 5) 4(x2 – 3x + 5) = 4x2 – 12x + 20 Check the answer. A common binomial factor can be factored out of certain expressions. Example: Factor the expression 5(x + 1) – y(x + 1). 5(x + 1) – y(x + 1) = (5 – y)(x + 1) (5 – y)(x + 1) = 5(x + 1) – y(x + 1) Check. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

A difference of squares can be factored using the formula a2 – b2 = (a + b)(a – b). Example: Factor x2 – 9y2. x2 – 9y2 = (x)2 – (3y)2 Write terms as perfect squares. = (x + 3y)(x – 3y) Use the formula. The same method can be used to factor any expression which can be written as a difference of squares. Example: Factor 4(x + 1)2 – 25y 4. 4(x + 1)2 – 25y 4 = (2(x + 1))2 – (5y2)2 = [(2(x + 1)) + (5y2)][(2(x + 1)) – (5y2)] = (2x + 2 + 5y2)(2x + 2 – 5y2) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Difference of Squares

Examples: 1. Factor 2xy + 3y – 4x – 6. Some polynomials can be factored by grouping terms to produce a common binomial factor. Examples: 1. Factor 2xy + 3y – 4x – 6. 2xy + 3y – 4x – 6 = (2xy + 3y) – (4x + 6) Group terms. = (2x + 3)y – (2x + 3)2 Factor each pair of terms. = (2x + 3)( y – 2) Factor out the common binomial. 2. Factor 2a2 + 3bc – 2ab – 3ac. 2a2 + 3bc – 2ab – 3ac = 2a2 – 2ab + 3bc – 3ac Rearrange terms. = (2a2 – 2ab) + (3bc – 3ac) Group terms. = 2a(a – b) + 3c(b – a) Factor. = 2a(a – b) – 3c(a – b) b – a = – (a – b). = (2a – 3c)(a – b) Factor. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Examples: Factor

Sum or Difference of Cubes A sum or difference of cubes can be factored using the formulas a3 + b3 = (a + b)(a2 – ab + b2) and a3 – b3 = (a – b)(a2 + ab + b2). Examples: 1. Factor x3 + 8y3. x3 + 8y3 = (x)3 + (2y)3 Write each term as a perfect cube. = (x + 2y)(x2 – x(2y) + (2y)2) Use the sum formula. = (x + 2y)(x2 – 2xy + 4y2) Simplify. 2. Factor 27a3 – y3z6. 27a3 – y3z6 = (3a)3 – ( yz2)3 Write each term as a perfect cube. = ((3a) – ( yz2))((3a)2 + (3a)( yz2) +( yz2)2) Use the difference formula. = (3a – yz2)(9a2 + 3ayz2 + y2z 4) Simplify. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Sum or Difference of Cubes

Therefore, x2 + 3x + 2 = (x + 1)(x + 2). To factor a trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example, x2 + 10x + 24 = (x + 4)(x + 6). One method of factoring trinomials is based on reversing the FOIL process. Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b. Example: Factor x2 + 3x + 2. x2 + 3x + 2 = (x + a)(x + b) F O I L = x2 + ax + bx + ab Apply FOIL to multiply the binomials. = x2 + (a + b) x + ab Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2. = x2 + (1 + 2) x + 1 · 2 Therefore, x2 + 3x + 2 = (x + 1)(x + 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Factor x2 + bx + c

It follows that both a and b are negative. Example: Factor x2 – 8x + 15. x2 – 8x + 15 = (x + a)(x + b) = x2 + (a + b)x + ab Therefore a + b = – 8 and ab = 15. It follows that both a and b are negative. Sum Negative Factors of 15 – 1, – 15 +15 – 3, – 5 – 8 x2 – 8x + 15 = (x – 3)(x – 5). Check: (x – 3)(x – 5) = x2 – 5x – 3x + 15 = x2 – 8x + 15. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Therefore a and b are two positive factors of 36 whose sum is 13. Example: Factor x2 + 13x + 36. x2 + 13x + 36 = (x + a)(x + b) = x2 + (a + b) x + ab Therefore a and b are two positive factors of 36 whose sum is 13. Sum Positive Factors of 36 1, 36 37 2, 18 20 15 3, 12 4, 9 13 6, 6 12 = (x + 4)(x + 9) x2 + 13x + 36 Check: (x + 4)(x + 9) = x2 + 9x + 4x + 36 = x2 + 13x + 36. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Example: Factor Completely A polynomial is factored completely when it is written as a product of factors that can not be factored further. Example: Factor 4x3 – 40x2 + 100x. 4x3 – 40x2 + 100x = 4x(x2) – 4x(10x) + 4x(25) The GCF is 4x. = 4x(x2 – 10x + 25) Use distributive property to factor out the GCF. = 4x(x – 5)(x – 5) Factor the trinomial. Check: 4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25) = 4x(x2 – 10x + 25) = 4x3 – 40x2 + 100x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor Completely

Factoring Polynomials of the Form ax2 + bx + c Factoring polynomials of the form ax2 + bx + c, (a  1) is usually done by trial and error. Example: Factor 2x2 + 5x + 3. 2x2 + 5x + 3 = (2x + a)(x + b) For some a and b. = 2x2 + (a + 2b)x + ab Therefore, a + 2b = 5 and ab = 3. a b a + 2b 1 3 7 – 1 – 3 – 7 3 1 5 2x2 + 5x + 3 = (2x + 3)(x + 1) – 3 – 1 – 5 Check: (2x + 3)(x + 1) = 2x2 + 2x + 3x + 3 = 2x2 + 5x + 3. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Factoring Polynomials of the Form ax2 + bx + c

This polynomial factors as (x + a)(4x + b) or (2x + a)(2x + b). Example: Factor 4x2 – 12x + 5. This polynomial factors as (x + a)(4x + b) or (2x + a)(2x + b). Since ab = + 5, a and b have the same sign. a = –1, b = – 5 or a = 1 and b = 5. The middle term –12x equals either (4a + b) x or (2a + 2b) x. Since a and b cannot both be positive, they must both be negative. Trial Factors Middle Term (x –1)(4x – 5) – 5x – 4x = – 9x Create trial factorizations with the forms above. (x –5)(4x – 1) – x – 20x = – 21x (2x –1)(2x – 5) – 10x – 2x = –12x 4x2 – 12x + 5 = (2x –1)(2x – 5) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Example: Factoring Trinomials Trinomials which are quadratic in form are factored like quadratic trinomials. Example: Factor 3x 4 + 28x2 + 9. 3x 4 + 28x2 + 9 = 3u2 + 28u + 9 Let u = x2. = (3u + 1)(u + 9) Factor. = (3x2 + 1)(x2 + 9) Replace u by x2. Many trinomials cannot be factored. Example: Factor x2 + 3x + 5. Let x2 + 3x + 5 = (x + a)(x + b) = x2 + (a + b) x + ab. Then a + b = 3 and ab = 5. This is impossible. The trinomial x2 + 3x + 5 cannot be factored. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factoring Trinomials