In algebra, factor by rewriting a polynomial as a product of lowerdegree polynomials


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1 Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Name: Date: Block: Factoring Review Factor: rewrite a number or expression as a product of primes; e.g. 6 = 2 3 In algebra, factor by rewriting a polynomial as a product of lowerdegree polynomials In the example above, (x + 1)(x 2) is the factored form of x 2  x 2 (multiply to verify!) We will look at 5 different factoring types (many thanks to Mrs. Donohue from TJHSST!) TYPE I Factoring: Factor Out GCF Monomials Find a common monomial in the polynomial (the GCF) and factor it out (perform reverse distribution) IMPORTANT!!! Always factor common factors out first (type I factoring); then factor the polynomial that remains, if necessary, using other factoring methods. IMPORTANT!!! Always verify your factored results by multiplying the factors to get the original polynomial Factor the following polynomials a) 6x x GCF = Factored: Verify by distributing: b) 12x + 42y GCF = Factored: c) 4x x 3 GCF = Factored: You Try Factor: a) 4m  2 b) 9m 3 3m 2 c) 3x x 6 Type II Factoring Sums and Differences of Perfect Squares and Cubes Remember to factor out any common factors first! Two Terms that are the Difference of Perfect Squares: a 2 b 2 o Difference of squares: a 2 b 2 = (a + b)(a b) Factor the binomials below a) y 2 16 b) 4x 225 c) x 4 16 d) x 81 e) 9x 6 y 8
2 Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Page 2 Two Terms that are Sum or Difference of Perfect Cubes: a 3 + b 3 or a 3 b 3 o Sum of cubes: a 3 + b 3 = (a + b)(a 2 ab + b 2 ) o Difference of cubes: a 3 b 3 = (a b) (a 2 + ab + b 2 ) o To remember the signs SOAP: Same Opposite Always Positive o The trinomial will not be factorable Factor the binomials below a) x 3 8 b) 27x c) x 3 y 6 64 d) 16y You Try: Factor the binomials below (factor out the GCF first if necessary!) a) 3x 227 b) 4x 216 c) 8x 2 50 d) x e) 8x 3 27y 3 f) x 15 + y 21 g) 2x 3 18x h) 4x i) 16x Type III Factoring Trinomials with Leading Coefficient 1 (form: x 2 + bx + c) Notice a special product: If it is of the form a 2 + 2ab + b 2, then its factored form is (a + b) 2. If it is of the form a 22ab + b 2, then its factored form is (a  b) 2. A quick test is to make a binomial of a and c and square it, and see if you get original polynomial, or use complete the square rules (c is (b/2) 2 ). a) x 2 + 6x + 9 b) x 2 10x + 25 c) x 2 + 4x + 4 For all other trinomials of the form x 2 + bx + c, you must ask yourself the question: What do you multiply to get the last number (c), and add to get the middle number (b)?
3 Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Page 3 a) x 2 + 5x + 6 b) x 2 6x + 8 c) x 2  x 2 d) x 103x 5 10 You Try Factor: a) x 2 + 4x + 3 b) x 2 11x + 24 c) x 2 + 6x  16 d) x 2 2x  24 e) x x 24 f) x 2 14x + 24 g) x 2 + 6xy 7y 2 h) x 4 + 4x 2 32 Type IV Factoring Factor by Grouping (Four Terms) If we are given a four term polynomial, we split the polynomial into two sets of two terms, and factor those sets using type I factoring. If we find a common polynomial, we use type I factoring again to factor it out. Factoring a common polynomial: Factor x(x 5) + 3(x  5) Notice there is a common polynomial of x 5. Use type I factoring to factor it out; we are left with x + 3. So the factored form is (x 5)(x + 3). a) 5x 2 (x 2) + 3(x 2) b) 7y(5 y) 3(y 5) c) 11x(x 8) + 3(8 x) Use this skill to factor a four term polynomial: o Factor the first two terms, then factor the second two terms. o Factor out the common polynomial.
4 Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Page 4 a) n 3 + 6n 2 + 5n + 30 b) m 3 + 7m 2 2m 14 c) 9x 3 7x + 9x 27 You try: Factor the expression a) 3y 2 (y 2) + 5(2 y) b) x 3 + 3x 2 + 5x + 15 c) x 2 y 2 + 4x + 4y d) x 3 + x 2 + x + 1 e) y 2 + y + xy + x f) x x 3x 2 (HINT: Rearrange terms in degree order!) Type V Factoring Factor ax 2 + bx + c We can factor polynomials of the form x 2 + bx + c (type III factoring). What do we do to factor polynomials of this form when the leading coefficient is not 1? IMPORTANT: Always factor out a GCF first; you may find that you really have a type III. Example: 2x 2 2x 4 = 2(x 2 x 2) = 2(x 2)(x + 1) Method 1: Guess and Check Factor 2x 2 7x + 3 Draw sets of parentheses: ( )( ) In this case, the first terms in each must be 2x and x (why?) and the signs must be negative (why?): (2x  )(x  ) The factors of 3 are 1 and 3; test by multiplying back to see what works o (2x 3)(x 1) o (2x 1)(x 3) Factors are (2x 1)(x 3) 2x 2 5x + 3 NOPE! 2x 2 7x + 3 YES!! Fairly easy to do when a and c are prime numbers; gets harder if they are not!
5 Algebra 2 Notes SOL AII.1 Factoring Polynomials Mrs. Grieser Page 5 Method 2: Factor by Grouping Method If you are not a good guesser, it can be hard sometimes to use the guess and check method. Factoring by grouping (type IV) can help us: Factor 15x x + 2 METHOD 1) Factor out GCF if there is one 2) Multiply a x c 3) What factors of ac add to b? 4) Split up middle term by factors found above 5) Apply grouping method (type IV) 6) Factor out polynomial 7) VERIFY (do not skip this step) 1) No common factors EXAMPLE 2) a = 15; c = 2; ac = 15 x 2 = 30 3) What factors of 30 add to 13? 10 and 3 4) Split up middle term: 15x x + 3x + 2 5) Group: 5x(3x + 2) + (3x + 2) 6) Factor out polynomial: (3x+2)(5x+1) 7) VERIFY: (3x + 2)(5x + 1) = 15x x + 2 NOTE: ALWAYS FACTOR OUT ANY GCF PRIOR TO USING METHOD ABOVE!! a) 6x 2 11x  10 b) 3x x  5 c) 4x x  14 You try: Factor the polynomials a) 3x 2 + 8x + 4 b) 4x 2 9x + 5 c) 2x 2 13x + 6 d) 4x x + 7 e) 4x x  3 f) 12x 2 x  6
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