MATH 102 College Algebra


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1 FACTORING Factoring polnomials ls is simpl the reverse process of the special product formulas. Thus, the reverse process of special product formulas will be used to factor polnomials. To factor polnomials will mean to epress it as a product of positive integral powers of distinct prime factors. TYPES OF FACTORING Tpe 1. Common Monomial Factor a + a = a( + ) = 3 (4 3). 10a 6 b 5 15a 4 b 6 0a 3 b 4 The common factor of 10, 15 and 0 is 5; For a 6, a 4, a 3, the common factor is a 3 For b 5, b 6 and b 4, the common factor is b 4 Therefore, the common monomial factor of the given polnomial is 5a 3 b 4 After getting the common monomial factor, divide the given polnomial b this to get the other factor. 10a 6 b 5 15a 4 b 6 0a 3 b 4 = 5a 3 b 4 (a 3 b + 3ab 6 + 4) Tpe. Difference of Two Squares The difference of two squares is equal to the product of the sum and difference of the square roots of the terms. = ( + ) ( ) 1. 9 = = (  3) ( + 3) = (4 ) (9 ) = (4 9 ) (4 + 9 ) = ( 3) ( 3) (4 + 9 ) 3. a + 9a = a ( 9) = a ( 3)( + 3) Tpe 3. Perfect Square Trinomial The square of an binomial is a perfect square trinomial where the first and the last terms are the square of the first and square of last term of the binomial and the other term is a plus (or minus) the product of the first and the product of the first and last term of the binomial. + + = ( + ) + = ( ) To check if the trinomial is a perfect square trinomial, two terms should be perfect squares the other term should be a plus or minus twice the product of the square roots of the other terms. 5
2 1. Find the factors of Check if the given is perfect square trinomial: Two terms are perfect squares, 4 and 5. The other term, 0 is the product of the square root of 4, which is and the square root of 5 which is 5. Therefore, = ( 5). 9a + 30ab + 5b = (3a + 5b) (get the common factor first) = ( ) = (10 11) EXERCISES: Factor the following polnomials: A a 5 b 16a 3 b a 3 9ab B a 6 16c 6 3. a b k 8 5. m 4 b k 1 8. (a + b) b n c n b n c n C. 1. m 6m + 9. k 34k +k z + 5z 4. 5d 80d v 4 4v w + 16w p 3 0p + 11p
3 Tpe 4. Sum and Difference of Two Cubes B long division, we could verif that = + and = = ( + )( + ) 3 3 = ( )( + + ) = (3 ) ( ). a = (a + 4)(a 4a + 16) 3. 15b 4 c bc 4 = bc(15b 3 c 3 ) = bc(5b c)(5b + 5bc + c ) Tpe 5. Other Trinomials (Trial & Error Method) Certain trinomials of the form + (a +b) + ab, can onl be factored b trial and error method because the sum of the products of the means and the etremes of the factors should be equal to the middle (linear) term of the given trinomial. + (a +b) + ab = ( + a) ( + b) ac + (ad + bc) + bd = (a + b)(c + d) 1. Factor 8. Find the value of a and b where + (a +b) + ab = ( + a) ( + b) = 8 ab =  8, thus a and b has opposite signs (a + b) = , the factors of 8 that will give a sum of are 4 and. The factors, therefore, are ( 4) ( + ). 8 = ( 4) ( + ) Find the value of a,b, c and d where ac + (ad + bc) + bd = (a + b)(c + d) = ac = 15 bd =  8 (ad + bc) = Possible factors for ac are 3 & 5, 3 & 5, 15 & 1, 15 & 1 Possible factors for bd are 4 &, 4 & , 8 & 1 and 8 & 1 From these pairs of factors, the pair that will give (ad + bc) = are 3 & 5 and  & 4 Therefore, = (3 )(5 + 4) Find the value of a,b, c and d where ac + (ad + bc) + bd = (a + b)(c + d) = ac = 6 bd =  15 (ad + bc) = 1 Possible factors for ac are 3 &, 6 & 1 Possible factors for bd are 3 & 5, 3 & 5, 15 & 1, 15 & 1 From these pairs of factors, The pair that will give (ad + bc) = 1 are 3 & and 5 & 3 Therefore, = (3 5)( + 3) 7
4 Tpe 7. Factoring B Grouping Sometimes proper grouping of terms is necessar to make the given polnomial factorable. After terms are grouped, a complicated epression ma be factored easil b appling Tpes 1 to 5 formulas. This tpe of factoring is usuall applied to algebraic epressions consisting of at least four terms. 1. 3(a b) + 4(a b) =(a b)(3 + 4) common factor. b + b + h + h = (b + b) + (h + h) grouping = b( + ) + h( + ) removal of common factor from each group = ( + )(b+ h) Another solution: = (b + h) + (b + h) = (b + h) + (b + h) = (b + h)( + ) 3. 3(a b) + 4(b a) = 3(a b) 4(a b) alteration of sign = (a b)(3 4) 4. z k + kw wz = (z k) (wz kw) = (z k) w(z k) = (z k)( w) 5. ab 3 3b 4a + 1 = (ab 3 3b ) (4a 1) = b (a 3) 4(a 3) = (a 3) (b 4) = (a 3) (b ) (b+) Tpe 8. Factoring B Addition or Subtraction of Suitable Terms This tpe of factoring is usuall applied to polnomials of degree 4 with two terms being perfect squares and both preceded b positive sign. Through addition or subtraction of suitable terms, the given will alwas lead to the difference of two squares. 1. v (a polnomial of degree 4, with two terms being perfect square and are both preceded positive sign.) to factor: we add + 4v and 4v, thus, the value of the polnomial is not changed because + 4v and 4v is equal to 0. v 4 + 4v + 4 4v = (v 4 + 4v + 4) 4v = (v + ) (v) (difference of two squares) = [(v + ) + v] [(v + ) v] = (v +v + )(v v + ). z 4 + 5z + 9 (z 4 and 9 are perfect squares) we add + z and z to make z 4 + 5z + 9 a perfect trinomial square z 4 + 5z + z + 9 z = (z 4 + 6z + 9) z = (z + 3) z = (z z)(z + 3 z) = (z + z + 3)(z z + 3) EXERCISES: Factor the following polnomials completel: A. Factoring b Grouping uv +3u + v rs + 4r + s ac ad + bc bd a ab + a b 7. c + 4cd 3c 6d
5 B. Factoring b Addition or Subtraction of Suitable Terms a 4 + 4a b 4 + 5b m 4 7m r r k 4 1k b b m 4 + 3m + 4 9
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