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Objective. 1.Factor quadratic trinomials of the form x 2 + bx + c. 2. Factor quadratic trinomials of the form x 2 + bx + c. 3. Factor four terms by grouping (front 2, back 2). Example 1A: Factoring by Using the GCF. Factor each polynomial. Check your answer. – 14 x – 12 x 2.
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Objective 1.Factor quadratic trinomials of the form x2 + bx + c. 2. Factor quadratic trinomials of the form x2 + bx + c. 3. Factor four terms by grouping (front 2, back 2)
Example 1A: Factoring by Using the GCF Factor each polynomial. Check your answer. –14x– 12x2 Check–2x(7 + 6x) Multiply to check your answer. –14x– 12x2 The product is the original polynomial.
Example 1B Factor each polynomial. Check your answer. 8x4 + 4x3– 2x2 8x4 = 2 2 2 x x x x 4x3 = 2 2 x x x Find the GCF. 2x2 = 2 x x 2 x x = 2x2 The GCF of 8x4, 4x3 and –2x2 is 2x2. Write terms as products using the GCF as a factor. 4x2(2x2) + 2x(2x2)–1(2x2) 2x2(4x2 + 2x– 1) Use the Distributive Property to factor out the GCF. Check 2x2(4x2 + 2x– 1) Multiply to check your answer. 8x4 + 4x3– 2x2 The product is the original polynomial.
Example 2 Factor each expression. a. 4s(s + 6) – 5(s + 6) The terms have a common binomial factor of (s + 6). 4s(s + 6)– 5(s + 6) (4s – 5)(s + 6) Factor out (s + 6). b. 7x(2x + 3) + (2x + 3) 7x(2x + 3) + (2x + 3) The terms have a common binomial factor of (2x + 3). 7x(2x + 3) + 1(2x + 3) (2x + 1) = 1(2x + 1) (2x + 3)(7x + 1) Factor out (2x + 3).
You may be able to factor a polynomial by grouping. When a polynomial has four terms, you can make two groups and factor out the GCF from each group.
Example 3A: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 6h4– 4h3 + 12h– 8 Group terms that have a common number or variable as a factor. (6h4– 4h3) + (12h– 8) 2h3(3h– 2) + 4(3h– 2) Factor out the GCF of each group. 2h3(3h– 2) + 4(3h– 2) (3h – 2) is another common factor. (3h– 2)(2h3 + 4) Factor out (3h – 2).
6h4– 4h3 + 12h– 8 Example 3A Continued Factor each polynomial by grouping. Check your answer. Check (3h– 2)(2h3 + 4) Multiply to check your solution. 3h(2h3) + 3h(4) – 2(2h3) – 2(4) 6h4 + 12h –4h3– 8 The product is the original polynomial.
Example 43B: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 5y4– 15y3 + y2– 3y (5y4– 15y3) + (y2– 3y) Group terms. Factor out the GCF of each group. 5y3(y – 3) + y(y– 3) (y – 3) is a common factor. 5y3(y– 3) + y(y– 3) (y– 3)(5y3 + y) Factor out (y – 3).
Example 4: Factoring with Opposites Factor 2x3– 12x2 + 18 – 3x 2x3– 12x2 + 18 – 3x (2x3– 12x2) + (18 – 3x) Group terms. 2x2(x– 6) + 3(6 –x) Factor out the GCF of each group. 2x2(x– 6) + 3(–1)(x– 6) Write (6 – x) as –1(x – 6). 2x2(x– 6) – 3(x– 6) Simplify. (x – 6) is a common factor. (x –6)(2x2– 3) Factor out (x – 6).
Lesson Quiz: Part 1 Factor each polynomial. Check your answer. 1. 16x + 20x3 2. 4m4 – 12m2 + 8m Factor each expression (by grouping). 3. 3y(2y + 3) – 5(2y + 3) 4x(4 + 5x2) 4m(m3 – 3m + 2) (2y + 3)(3y – 5) 4. 2x3 + x2 – 6x – 3 5. 7p4 – 2p3 + 63p – 18 (2x + 1)(x2 – 3) (7p – 2)(p3 + 9)
Remember! When you multiply two binomials, multiply: First terms Outer terms Inner terms Last terms Notice that when you multiply (x + 2)(x + 5), the constant term in the trinomial is the product of the constants in the binomials. (x +2)(x +5) = x2 + 7x +10
x2 12 3x The guess and check method is usually not the most efficient method of factoring a trinomial. Look at the product of (x + 3) and (x + 4). (x + 3)(x +4) =x2+7x+12 4x The coefficient of the middle term is the sum of 3 and 4. The third term is the product of 3 and 4.
(x + )(x + ) Factors of 5 Sum 1 and 5 6 = x2 + 6x + 5 Example 1: Factoring x2 + bx + c Factor each trinomial. Check your answer. x2 + 6x + 5 b = 6 and c = 5; look for factors of 5 whose sum is 6. The factors needed are 1 and 5. (x + 1)(x + 5) Check(x + 1)(x + 5) = x2 + x + 5x + 5 Use the FOIL method. The product is the original polynomial.
(x + )(x + ) Factors of –15 Sum –1 and –15 –16 –3 and –5 –8 = x2 – 8x + 15 Example 2: Factoring x2 + bx + c Factor each trinomial. Check your answer. x2– 8x + 15 b = –8 and c = 15; look for factors of 15 whose sum is –8. The factors needed are –3 and –5 . (x–3)(x–5) Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method. The product is the original polynomial.
(x + )(x + ) Factors of –20 Sum –1 and 20 19 –2 and 10 8 –4 and 5 1 Example 3: Factoring x2 + bx + c Factor each trinomial. x2 + x– 20 b = 1 and c = –20; look for factors of –20 whose sum is 1. The factor with the greater absolute value is positive. The factors needed are +5 and –4. (x– 4)(x + 5)
(x + )(x + ) Factors of –15 Sum –1 and 15 14 –3 and 5 2 Example 4 Factor each trinomial. Check your answer. x2 + 2x– 15 b = 2 and c = –15; look for factors of –15 whose sum is 2. The factor with the greater absolute value is positive. The factors needed are –3 and 5. (x– 3)(x + 5)
Lesson Quiz: Part I Factor each trinomial. 1.x2 – 11x + 30 2.x2 + 10x + 9 3.x2 – 6x – 27 4.x2 + 14x – 32 (x – 5)(x – 6) (x + 1)(x + 9) (x– 9)(x + 3) (x + 16)(x – 2)