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Chabot Mathematics. §5.4 Factor TriNomials. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. MTH 55. 5.3. Review §. Any QUESTIONS About §5.3 → Factoring by GCF and/or Grouping Any QUESTIONS About HomeWork §5.3 → HW-13. Factor (+1) · x 2 + bx + c.

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## Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

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**Chabot Mathematics**§5.4 FactorTriNomials Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu**MTH 55**5.3 Review § • Any QUESTIONS About • §5.3 → Factoring by GCF and/or Grouping • Any QUESTIONS About HomeWork • §5.3 → HW-13**Factor (+1)·x2 + bx + c**• Recall the FOIL method of multiplying two binomials: F O I L (x + 2)(x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10**Factor (+1)·x2 + bx + c**• To factor x2 + 7x + 10, think of FOIL: The first term, x2, is the product of the First terms of two binomial factors, so the first term in each binomial must be x. The challenge is to find two numbers p and q such that x2 + 7x + 10 = (x + p)(x + q) = x2 + qx + px + pq**Factor (+1)·x2 + bx + c**• Need to find two numbers p & q such that x2 + 7x + 10 = (x + p)(x + q) = x2 + qx + px + pq • Thus the numbers p and q must be selected so that their • PRODUCT is 10 • SUM is 7 • The Factor Pairs for 10 [and their sums] • 1·10 [11]; (−1)·(−10) [−11]; 2·5 [7]; (−2)·(−5) [−7];**Factor (+1)·x2 + bx + c**• In this case, Examination of the “c” term factor-pairs revealed the desired numbers p = 2 & q = 5 • Thus the factorization (x + 2)(x + 5), or (x + 5)(x + 2).**Example FOIL Factoring**• Multiplying binomials uses the FOIL method, and factoring usestheFOILmethod backwards Product of x and x is x2. F L Product of 5 and –7 is –35. Sum of the product of outer and inner terms O I**Factor x2 + bx + c for Positive c**• When the constant term of a trinomial is positive, look for two numbers with the same sign. The sign is that of the middle term: x2– 7x + 10 = (x – 2)(x – 5); x2 + 7x + 10 = (x + 2)(x + 5);**SOLUTION:Think of FOIL in reverse:**(x + )(x + ) We need a constant term that has a product of 12 and a SUM of 7. We list some pairs of numbers that multiply to 12 Example Factor x2 + 7x + 12**Example Factor x2 + 7x + 12**• Since 3 4 = 12 and 3 + 4 = 7,the factorization of x2 + 7x + 12 is (x + 3)(x + 4). • To check we simply multiply the two binomials. • CHECK by FOIL: (x + 3)(x + 4) = x2 + 4x + 3x + 12 = x2 + 7x + 12 **Example Factor y2– 8y + 15**• SOLUTION: Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 15 in which both factors are negative. Their SUM must be −8. Sum of −8 y2− 8y + 15 = (y −3)(y– 5)**Factor x2 + bx + c for Negative c**• When the constant term of a trinomial is negative, look for two numbers whose product is negative. One must be positive and the other negative: x2– 4x – 21 = (x + 3)(x – 7); x2 + 4x – 21 = (x – 3)(x + 7). • Select the two numbers so that the number with the LARGER absolute value has the SAME SIGN as b, the coefficient of the middle term**SOLUTION: The constant term must be expressed as the product**of negative & positive numbers. Since the sum of the two numbers must be negative, the negative number must have the greater absolute value. Example Factor x2– 5x– 24 x2− 5x− 24 =(x + 3)(x– 8)**SOLUTION:**Rewrite the trinomialt2 + 4t− 32. We need one positive and one negative factor. The sum must be 4, so the positive factor must have the larger absolute value Example Factor t2– 32 + 4t t2 + 4t− 32= (t + 8)(t− 4)**Example Two Variables**• Factor: a2 + ab− 30b2 • SOLUTION • We need the factors of a2 & 30b2 that when added equal ab. • Those factors are a, and −5b & 6b. a2 + ab− 30b2 = (a− 5b)(a + 6b)**Prime Polynomials**• A polynomial that canNOT be factored is considered prime. • Example: x2−x + 7 • Often factoring requires two or more steps. Remember, when told to factor, we should factor completely. This means the final factorization should contain only prime polynomials.**Example Factor 2x3−24x2+72x**• SOLUTIONAlways look first for a common factor. In this case factor out 2x: 2x(x2− 12x + 36) • Since the constant term is positive and the coefficient of the middle term is negative, we look for the factorization of 36 in which both factors are negative. • Their SUM must be −12.**The factorization of(x2– 12x + 36) is (x– 6)(x – 6) or**(x– 6)2 Example Factor 2x3–24x2+72x • The factorization of 2x3– 24x2 + 72x is 2x(x– 6)2or 2x(x – 6)(x – 6) 2x3– 242x– 72x = 2x(x– 6 )(x– 6)**To Factor (+1)·x2 + bx + c**• Distribute out Common Factors • Find a pair of factors that have c as their product and b as their sum. • If c is positive, its factors will have the same sign as b. • If c is negative, one factor will be positive and the other will be negative. Select the factors such that the factor with the larger absolute value has the same sign as b. • CHECK by MULTIPLYING**Factoring When: LeadCoeff ≠ 1**• Factoring Trinomials of the Type ax2 + bx + c • Factoring with FOIL • The Grouping Method**Example Factor 3x2– 14x– 5**• First, check for a common factor, or GCF for all Terms. There is none other than 1 or −1. • Find the First terms whose product is 3x2. The only possibilities are 3x and x: (3x + )(x + ) • Find the Last terms whose product is −5.Possibilities are (−5)(1) & (5)(−1) • Important!: Since the First terms are not identical, we must also consider the above factors in reverse order: (1)(−5), & (−1)(5).**Example Factor 3x2– 14x– 5**• Knowing that the First and Last products will check, inspect the Outer and Inner products resulting from steps (2) and (3) Look for the combination in which the sum of the products is the middle term. (3x– 5)(x + 1) = 3x2+ 3x– 5x– 5 = 3x2– 2x– 5 (3x– 1)(x + 5) = 3x2+ 15x–x– 5 = 3x2+ 14x– 5 Wrong middle term Wrong middle term**Example Factor 3x2– 14x– 5**• Keep Trying the factors of −5. (3x + 5)(x– 1) = 3x2– 3x + 5x– 5 = 3x2+ 2x– 5 (3x + 1)(x– 5) = 3x2– 15x + x– 5 = 3x2– 14x– 5 Wrong middle term Correct middle term! • Thus 3x2– 14x– 5 = (3x + 1)(x– 5)**LdCoeff ≠ 1 Factorization Notes**• Reversing the signs in the binomials reverses the sign of the middle term • Organize your work so that you can keep track of which possibilities you have checked. • Remember to include the largest common factor - if there is one - in the final factorization. • ALWAYS CHECK by multiplying**Example Factor 14x + 5– 3x2**• SOLUTION: • It is an important problem-solving strategy to find a way to make problems look like problems we already know how to solve. Rewrite the equation in descending order. 14x + 5– 3x2 = – 3x2+ 14x + 5**Example Factor 14x + 5− 3x2**• Starting with −3x2 + 14x + 5 • Factor out the –1: −3x2 + 14x + 5 = −1(3x2− 14x− 5) = −1(3x + 1)(x− 5) • The factorization of 14x + 5− 3x2 is −1(3x + 1)(x− 5). or (−3x − 1)(x− 5) or (3x + 1)(−x + 5)**Example 2Vars: 6x2−xy− 12y2**• SOLUTION: No common factors exist, we examine the first term, 6x2. There are two possibilities: (2x + )(3x + ) or (6x + )(x + ). • The last term −12y2, has pairs of factors: 12y, −y 6y, −2y 4y, −3y and −12y, y −6y, 2y −4y, 3y as well as each pairing reversed.**Example 2Vars: 6x2−xy− 12y2**• SOLUTION: • Some trials such as (2x– 6y)(3x + 2y) and (6x + 4y)(x– 3y), cannot be correct because (2x– 6y) and (6x + 4y) contain a common factor, 2. • Trial (2x + 3y)(3x− 4y) • Product 6x2− 8xy + 9xy− 12y2 = 6x2 + xy− 12y2**Example 2Vars: 6x2−xy− 12y2**• SOLUTION: the Trial (2x + 3y)(3x− 4y) incorrect, but only because of the sign of the middle term. • To correctly factor, simply change the signs in the binomials. • Trial (2x− 3y)(3x + 4y) • Product 6x2 + 8xy− 9xy− 12y2 = 6x2−xy− 12y2 The factorization: (2x− 3y)(3x + 4y)**Example 18m2 – 19mn – 12n2**• SOLUTION • There are no common factors. • Factor the first term, 18m2 and get the following possibilities: 18mm, 9m2m, and 6m3m. • Factor the last term, −12n2, which is negative. The possibilities are: (−12n)(n), (−n)(12n), (−2n)(6n), (6n)(−2n), (−4n)(3n) or (−3n)(4n)**Example 18m2 – 19mn – 12n2**• Look for combinations of factors such that the sum of the outside and the inside products is the middle term, (−19mn). • (9m + n)(2m− 12n) = 18m2− 106mn− 12n2 • (9m− 12n)(2m + n) = 18m2− 15mn− 12n2 • (9m− 3n)(2m + 4n) = 18m2 + 30mn− 12n2 • (9m + 4n)(2m – 3n) = 18m2− 19mn− 12n2 • Thus ANS → (9m + 4n)(2m− 3n) correct middle term**Factoring ax2+bx+c by Grouping**• Factor out the largest common factor, if one exists. • Multiply the leading coefficient a and the constant c; i.e., form the a•c product • Find a pair of factors of a•c whose sum is b. • Rewrite the middle term, bx, as a sum or difference using the factors found in step (3). • Factor by grouping. • Include any common factor from step (1) and check by multiplying.**Example Factor 4x2– 5x– 6**• SOLUTION • First, we note that there is no common factor (other than 1 or −1). • We multiply the leading coefficient, 4 and the constant, −6: (4)(−6) = −24. • We next look for the factorization of −24 in which the sum of the factors is the coefficient of the middle term, −5.**Example Factor 4x2– 5x– 6**We would normally stop listing pairs of factors once we have found the one we need**Example Factor 4x2– 5x– 6**• Next, we express the middle term as a sum or difference using the factors found in step (3): −5x = −8x + 3x. 5. We now factor by grouping as follows: 4x2− 5x− 6 = [4x2− 8x] + [3x −6] = 4x(x− 2) + 3(x −2) = (x− 2)(4x + 3)**Example Factor 4x2– 5x– 6**• CHECK by FOIL: (x− 2)(4x + 3) = 4x2 + 3x − 8x− 6 = 4x2− 5x− 6 • The factorization of 4x2− 5x− 6 is (x− 2)(4x + 3).**Example: Factor 8x3 + 10x2– 12x**• SOLUTION • Factor out the Greatest Common Factor (GCF), 2x: 8x3 + 10x2− 12x = 2x(4x2 + 5x− 6) 2. To factor 4x2 + 5x− 6 by grouping, we multiply the leading coefficient, 4 and the constant term (−6): 4(−6) = −24.**Example: Factor 8x3 + 10x2– 12x**• next look for pairs of factors of −24 whose sum is 5. • We then rewrite the 5x in 4x2 + 5x− 6 using: 5x = −3x + 8x**Example: Factor 8x3 + 10x2– 12x**• Next, factor by grouping: 4x2 + 5x− 6 = [4x2− 3x] + [8x −6] = x(4x− 3) + 2(4x −3) = (x + 2)(4x− 3) • The factorization of the original trinomial 8x3 + 10x2− 12x is 2x(x + 2)(4x− 3)**Factoring by Substitution**• Some times substituting a single varible for a (complicated) expression reveals an easily factored PolyNomial • Example factor p2q2 + 7pq + 6 • SOLUTION • Rewrite using Product-to-Power Exponent rule → (pq)2 + 7(pq) + 6 • Now engage a substitution**Factoring by Substitution**• factor p2q2 + 7pq + 6 = (pq)2 + 7(pq) + 6 • Now to engage a substitution LET u = pq • Replace in the Expression pq with u u2 + 7u + 6 • The Expression in u is easily FOIL-factored u2 + 7u + 6 = (u + 6)(u + 1)**Factoring by Substitution**• Now BACK Substitute u = pq (u)2 + 7 (u) + 6 = (u + 6)(u + 1) (pq)2 + 7(pq) + 6 = (pq + 6)(pq + 1) p2q2 + 7pq + 6 = (pq + 6)(pq + 1)**WhiteBoard Work**• Problems From §5.4 Exercise Set • 30, 44, 66, 82, 92 • Shaded AreaEquals**All Done for Today**F.O.I.L.Factoring**Chabot Mathematics**Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –**Graph y = |x|**• Make T-table

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