UNIT 1PolynomialsLesson 5 of 61 Adding, Subtracting and Multiplying Polynomials2 Special Products3 Factoring: GCF and Grouping4 Factoring x² + bx + c5 Factoring ax² + bx + c6 Factoring Special Cases
Factoring ax² + bx + c
When x² has a number in front, split the middle term into two parts. Then factor by grouping.
What you'll be able to do
- Factor ax² + bx + c with the ac method.
- Take out a GCF first, even a negative one.
- Factor trinomials with two letters.
- Tell when a trinomial is prime.
The big idea
The ac method. Multiply a and c. Find two numbers that multiply to ac and add to b.
Split the middle. Write bx as two terms with those numbers. Then you have four terms.
Group. Factor the first pair and the last pair. Both pairs leave the same binomial. Take it out.
GCF first. 6x² + 3x − 18 = 3(2x² + x − 6). Factor what is left.
A negative first term. Take out −1 first. −2x² + x + 6 = −(2x² − x − 6).
Prime. If no pair multiplies to ac and adds to b, the trinomial is prime.
Check by multiplying. Multiply your factors. You should get back the trinomial.
Words you need
- Factor
- A part you multiply. To factor means to write something as a product.
- Product
- The answer when you multiply.
- Trinomial
- A polynomial with 3 terms, like 3x² + 11x + 6.
- Binomial
- A polynomial with 2 terms, like 3x + 2.
- Polynomial
- A sum of terms, like 12x³ + 56x² + 32x.
- Term
- One part of a polynomial, like 11x or 6.
- Coefficient
- The number in front of a letter. In 3x², it is 3.
- Leading coefficient
- The coefficient of the term with the highest exponent. In 3x² + 11x + 6, it is 3.
- Constant
- A term with no letter, like 6.
- ac method
- Multiply a and c, find two numbers with that product that add to b, split the middle term, then group.
- Grouping
- Factoring four terms in two pairs, then taking out the binomial both pairs share.
- Greatest common factor
- The GCF. The biggest factor that every term shares.
- Prime trinomial
- A trinomial that can't be factored with whole numbers.
- Integer
- A whole number or its negative, like −3, 0 or 7.
- Exponent
- The small raised number. In x², it is 2.
- Power
- A base with an exponent, like x².
- Base
- The bottom of a shape, like the bottom of a box. In a power like 3², the base is 3.
- Variable
- A letter that stands for a number, like x.
- Expression
- Numbers and letters joined by +, −, × or ÷. It has no = sign.
- Simplify
- Write it in a shorter form with the same value.
- Zero
- An input value that makes a function equal 0.
The ac method, step by step
| Step | Result |
|---|---|
| Start | 2x² + 7x + 3 |
| Multiply a and c | 2 × 3 = 6 |
| Find two numbers | 6 and 1: they multiply to 6 and add to 7 |
| Split the middle term | 2x² + 6x + x + 3 |
| Group | 2x(x + 3) + 1(x + 3) |
| Take out the shared binomial | (2x + 1)(x + 3) |
Worked examples
Example 1
Factor 3x² + 11x + 6.
- ac = 3 × 6 = 18. The numbers 9 and 2 multiply to 18 and add to 11.
- Split the middle term: 3x² + 9x + 2x + 6.
- Group: 3x(x + 3) + 2(x + 3).
- Take out (x + 3): (3x + 2)(x + 3).
Answer: (3x + 2)(x + 3)
Example 2
Factor 4x² − 4x − 3.
- ac = 4 × (−3) = −12. The numbers −6 and 2 multiply to −12 and add to −4.
- Split the middle term: 4x² − 6x + 2x − 3.
- Group: 2x(2x − 3) + 1(2x − 3).
- Take out (2x − 3): (2x + 1)(2x − 3).
Answer: (2x + 1)(2x − 3)
Example 3
Factor 6x² + 3x − 18 completely.
- Every term shares 3. Take it out: 3(2x² + x − 6).
- For 2x² + x − 6, ac = −12. The numbers 4 and −3 multiply to −12 and add to 1.
- Split and group: 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2).
- So 6x² + 3x − 18 = 3(2x − 3)(x + 2).
Answer: 3(2x − 3)(x + 2)
Example 4
Factor −2x² + x + 6.
- The first term is negative. Take out −1: −(2x² − x − 6).
- ac = −12. The numbers −4 and 3 multiply to −12 and add to −1.
- Split and group: 2x² − 4x + 3x − 6 = 2x(x − 2) + 3(x − 2) = (2x + 3)(x − 2).
- So −2x² + x + 6 = −(2x + 3)(x − 2).
Answer: −(2x + 3)(x − 2)
Common mistakes
Putting the ac numbers straight into the binomials. For 2x² + 7x + 3, the numbers are 6 and 1. But (2x + 6)(x + 1) is wrong. Split the middle term and group.
Skipping the GCF. (4x − 4)(2x − 3) is not done, because 4x − 4 still shares 4. Write 4(x − 1)(2x − 3).
A sign error after a minus. In 3x² + 6x − 4x − 8, take −4 out of the last pair: −4(x + 2), not −4(x − 2).
Dropping the 1. In 2x(2x − 3) + (2x − 3), the second group is 1(2x − 3). The answer is (2x + 1)(2x − 3).
Practice: 20 questions
12 of 20 are word problems. For a word problem, sketch it first and mark what you know.
Tier 1 Foundations Use the rule directly
- Factor 2x² + 11x + 12 and 3y² − 10y + 8. Check both answers by multiplying the binomials.
- A rectangular solar panel has an area of 3x² + 14x + 15 square feet, where x is greater than 1. Its length is greater than its width. Write binomials for the length and the width of the panel in feet, and find both dimensions in feet when x = 4. Word problem
- Brisa factors 2x² + 5x − 12. She finds two numbers that multiply to −24 and add to 5, which are 8 and −3, and then she writes (2x + 8)(x − 3) as her final answer. Explain how multiplying her factors shows her mistake, and write the correct factored form.
- Two students factor the area of a rectangular courtyard, 6x² + 7x − 3 square meters, by splitting the middle term in opposite orders. Tariq writes 6x² + 9x − 2x − 3 and groups the first two terms and the last two terms. Melina writes 6x² − 2x + 9x − 3 and groups the first two terms and the last two terms. Finish both students' groupings to decide who is right, or whether both ways work. What are the dimensions of the courtyard in meters when x = 5? Word problem
- A rectangular loading dock has an area of 4x² − 4x − 15 square meters, where x is greater than 3. Write binomials for the length and the width of the dock in meters, with the length greater than the width. Find the perimeter of the dock in meters when x = 6. Word problem
Tier 2 Intermediate One more step to think through
- A rectangular wooden planter has a volume of 12x³ + 56x² + 32x cubic inches. Write the volume polynomial in completely factored form to find three expressions for the dimensions of the planter in inches. Find the three dimensions in inches when x = 3. Word problem
- Factor −3x² + 13x − 4 completely so that the x terms inside the binomials have positive coefficients, and factor 5a² + 7ab − 6b² completely.
- A rectangular mural has an area of 12x² − 18x − 12 square feet, where x is greater than 2. Eamon factors the area as (4x + 2)(3x − 6) and marks his work complete. Explain why his answer is not completely factored, write the completely factored form, and find the area of the mural in square feet when x = 4. Word problem
- Compare the two trinomials 2x² + 7x + 6 and 2x² + 5x + 6. Which of them can be factored with integers? Factor any that can, and explain how the factor pairs of 12 show which one is prime.
- A rectangular stone patio has an area of 6x² + 19xy + 10y² square feet, where x and y are positive numbers. Write binomials for the dimensions of the patio in feet, and find the length and the width of the patio in feet when x = 4 and y = 3. Word problem
Tier 3 Advanced Several steps - find the hidden structure
- Find every positive whole number k that makes 3x² + kx + 4 factor into two binomials with whole numbers, and write the factored form for each value of k.
- A toy rocket launched from a platform has a height, in feet above the ground after t seconds, modeled by −16t² + 56t + 32. Write this polynomial in completely factored form with a negative leading factor. What is the height of the rocket in feet at t = 3 seconds? Word problem
- A rectangular swimming pool measures 2x + 3 meters long by x + 2 meters wide. A concrete deck around the pool adds 4x² + 22x + 29 square meters of area, so the combined area of the pool and the deck forms a larger rectangle whose length and width are binomials with whole-number coefficients. Find the polynomial for the combined area in standard form, and factor it to find the outer length and outer width in meters. Word problem
- Soraya factors 6x² − 11x − 10. She finds two numbers that multiply to −60 and add to −11, which are −15 and 4. She rewrites the trinomial as 6x² + 4x − 15x − 10, groups it as (6x² + 4x) − (15x − 10), and factors out the greatest common factor of each group to get 2x(3x + 2) − 5(3x − 2). Because 3x + 2 and 3x − 2 do not match, she claims the trinomial is prime. Find her mistake and write the correct factored form.
- Two rectangular garden plots sit end to end and share one side of the exact same binomial length, where x is greater than 4. Plot A has an area of 6x² + 5x − 6 square feet. Plot B has an area of 8x² + 2x − 15 square feet. Factor both area trinomials to find the binomial length of their shared side in feet. Which plot is longer along its other dimension for all x > 4, and by how many feet in terms of x? Word problem
Tier 4 Challenge Stretch problems, some above grade level
- A rectangular solar array has an area of 3x² + kx − 5 square meters, where x is greater than 6 and k is an integer. The length and the width are binomials with positive leading x coefficients and integer constants. Find every possible integer value of k, positive or negative, write the factored dimensions for each one, and find the largest possible perimeter of the solar array in meters in terms of x. Word problem
- Two students factor −6x² + 7x + 3. Ronan factors out −1 first and writes −(2x − 3)(3x + 1). Freja writes (3 − 2x)(3x + 1) with no negative sign in front of the parentheses. Ronan says Freja's answer must be wrong because the original leading coefficient is negative, so a minus sign must sit in front of the factored answer. Explain why Ronan's claim is mistaken, show how the two answers are related, and verify both by multiplying.
- Two rectangular shipping containers both have a height of x feet, where x is greater than 3, and they share one base dimension of the exact same binomial length. Container A has a volume of 12x³ + 8x² − 15x cubic feet. Container B has a volume of 12x³ + 4x² − 21x cubic feet. Factor both volume polynomials completely to find the shared base dimension and the other base dimension of each container in feet. Which container has the greater base perimeter for all x > 3, and by how many feet? Word problem
- A rectangular plaza is built from two square sections of side 2x + 1 meters, seven rectangular strips measuring 2x + 1 meters by 1 meter, and a corner section of area 6 square meters, giving a total area of 2(2x + 1)² + 7(2x + 1) + 6 square meters. Write the total area in completely factored form as the product of two simplified binomials for the overall length and width of the plaza in meters, with the length greater than the width. Word problem
- A zero of a polynomial function is an input value that makes the output equal 0, and a product of factors equals 0 when any factor equals 0. Consider the function y = −16t³ + 28t² − 6t. Write it in completely factored form with a negative leading factor, and find all three zeroes. Between the two positive zeroes, y is positive. How long is that interval? Above grade level
Answers
Tier 1 Foundations
- (2x + 3)(x + 4) and (3y − 4)(y − 2). For 2x² + 11x + 12, find two numbers whose product is 2 × 12 = 24 and whose sum is 11, which are 3 and 8. Split the middle term and group: 2x² + 8x + 3x + 12 = 2x(x + 4) + 3(x + 4) = (2x + 3)(x + 4). For 3y² − 10y + 8, find two numbers whose product is 3 × 8 = 24 and whose sum is −10, which are −6 and −4. Split the middle term and group: 3y² − 6y − 4y + 8 = 3y(y − 2) − 4(y − 2) = (3y − 4)(y − 2). Multiply to check: (2x + 3)(x + 4) = 2x² + 8x + 3x + 12 = 2x² + 11x + 12, and (3y − 4)(y − 2) = 3y² − 6y − 4y + 8 = 3y² − 10y + 8.
- Length is 3x + 5 feet and width is x + 3 feet; when x = 4, the length is 17 feet and the width is 7 feet. Multiply the leading coefficient and constant term: 3 × 15 = 45. The numbers 9 and 5 multiply to 45 and add to 14. Rewrite the middle term and factor by grouping: 3x² + 9x + 5x + 15 = 3x(x + 3) + 5(x + 3) = (3x + 5)(x + 3). For x > 1, 3x + 5 is greater than x + 3, so the length is 3x + 5 feet and the width is x + 3 feet. When x = 4, the length is 3(4) + 5 = 17 feet and the width is 4 + 3 = 7 feet.
- Multiplying (2x + 8)(x − 3) gives 2x² + 2x − 24 instead of 2x² + 5x − 12, and 2x + 8 still has a common factor of 2; the correct factored form is (2x − 3)(x + 4). Check Brisa's answer by multiplying: (2x + 8)(x − 3) = 2x² − 6x + 8x − 24 = 2x² + 2x − 24, which does not match 2x² + 5x − 12. Split the middle term 5x into 8x − 3x: 2x² + 8x − 3x − 12. Factor by grouping: 2x(x + 4) − 3(x + 4) = (2x − 3)(x + 4).
- Both students are right and get the same factors in reverse order, (2x + 3)(3x − 1) and (3x − 1)(2x + 3); when x = 5, the dimensions are 13 meters and 14 meters. Finish Tariq's grouping: 3x(2x + 3) − 1(2x + 3) = (2x + 3)(3x − 1). Finish Melina's grouping: 2x(3x − 1) + 3(3x − 1) = (3x − 1)(2x + 3). By the commutative property of multiplication, both factored forms are identical. Substitute x = 5: 2(5) + 3 = 13 meters and 3(5) − 1 = 14 meters.
- Length is 2x + 3 meters and width is 2x − 5 meters; when x = 6, the perimeter is 44 meters. Multiply 4 × (−15) = −60. The numbers −10 and 6 multiply to −60 and add to −4. Split the middle term and group: 4x² − 10x + 6x − 15 = 2x(2x − 5) + 3(2x − 5) = (2x + 3)(2x − 5). At x = 6, the length is 2(6) + 3 = 15 meters, the width is 2(6) − 5 = 7 meters, and the perimeter is 2(15 + 7) = 44 meters.
Tier 2 Intermediate
- 4x(3x + 2)(x + 4); when x = 3, the dimensions are 12 inches, 11 inches, and 7 inches. Factor out the greatest common factor 4x from all three terms: 4x(3x² + 14x + 8). For 3x² + 14x + 8, the numbers 12 and 2 multiply to 24 and add to 14: 3x² + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (3x + 2)(x + 4), giving 4x(3x + 2)(x + 4). At x = 3, the three dimensions are 4(3) = 12 inches, 3(3) + 2 = 11 inches, and 3 + 4 = 7 inches.
- −(3x − 1)(x − 4) and (5a − 3b)(a + 2b). Factor −1 out of −3x² + 13x − 4 to get −(3x² − 13x + 4). The numbers −12 and −1 multiply to 12 and add to −13: 3x² − 12x − x + 4 = 3x(x − 4) − 1(x − 4) = (3x − 1)(x − 4), giving −(3x − 1)(x − 4). For 5a² + 7ab − 6b², the numbers 10 and −3 multiply to −30 and add to 7: 5a² + 10ab − 3ab − 6b² = 5a(a + 2b) − 3b(a + 2b) = (5a − 3b)(a + 2b).
- Both binomials still contain common factors because 4x + 2 has a factor of 2 and 3x − 6 has a factor of 3; the completely factored form is 6(2x + 1)(x − 2), and the area when x = 4 is 108 square feet. Factor 2 out of 4x + 2 to get 2(2x + 1), and factor 3 out of 3x − 6 to get 3(x − 2). Multiply the constant factors 2 × 3 = 6 to get the completely factored form 6(2x + 1)(x − 2). At x = 4, the area is 6(2(4) + 1)(4 − 2) = 6(9)(2) = 108 square feet.
- 2x² + 7x + 6 factors as (2x + 3)(x + 2); 2x² + 5x + 6 is prime because the positive integer factor pairs of 2 × 6 = 12 have sums of 13, 8, and 7, so no pair adds to 5. In both trinomials, the product of the first and last coefficients is 2 × 6 = 12. List all positive integer factor pairs of 12 and their sums: 1 and 12 add to 13, 2 and 6 add to 8, and 3 and 4 add to 7. Because 3 + 4 = 7, 2x² + 7x + 6 factors as 2x² + 4x + 3x + 6 = (2x + 3)(x + 2). Because no pair adds to 5, 2x² + 5x + 6 is prime.
- The binomial dimensions are 2x + 5y feet and 3x + 2y feet; when x = 4 and y = 3, the length is 23 feet and the width is 18 feet. Multiply 6 × 10 = 60. The numbers 15 and 4 multiply to 60 and add to 19. Split the middle term and group: 6x² + 15xy + 4xy + 10y² = 3x(2x + 5y) + 2y(2x + 5y) = (2x + 5y)(3x + 2y). At x = 4 and y = 3, the dimensions are 2(4) + 5(3) = 23 feet and 3(4) + 2(3) = 18 feet.
Tier 3 Advanced
- k = 13 gives (3x + 1)(x + 4); k = 8 gives (3x + 2)(x + 2); k = 7 gives (3x + 4)(x + 1). Multiply the leading coefficient and constant term: 3 × 4 = 12. List all pairs of positive whole numbers whose product is 12, and add each pair to find k: 1 + 12 = 13, 2 + 6 = 8, and 3 + 4 = 7. Group for each value of k to get (3x + 1)(x + 4) for k = 13, (3x + 2)(x + 2) for k = 8, and (3x + 4)(x + 1) for k = 7.
- −8(2t + 1)(t − 4); 56 feet at t = 3 seconds. Factor out the negative greatest common factor −8 from all three terms: −8(2t² − 7t − 4). For 2t² − 7t − 4, the numbers −8 and 1 multiply to −8 and add to −7: 2t² − 8t + t − 4 = 2t(t − 4) + 1(t − 4) = (2t + 1)(t − 4), giving −8(2t + 1)(t − 4). At t = 3, the height is −8(2(3) + 1)(3 − 4) = −8(7)(−1) = 56 feet.
- Combined area is 6x² + 29x + 35 square meters; the outer dimensions are 3x + 7 meters and 2x + 5 meters. Multiply the pool dimensions to find the pool area: (2x + 3)(x + 2) = 2x² + 7x + 6 square meters. Add the deck area to get the combined area: (2x² + 7x + 6) + (4x² + 22x + 29) = 6x² + 29x + 35 square meters. For 6x² + 29x + 35, the numbers 15 and 14 multiply to 210 and add to 29: 6x² + 15x + 14x + 35 = 3x(2x + 5) + 7(2x + 5) = (3x + 7)(2x + 5).
- When Soraya placed parentheses after the minus sign around −15x − 10, she failed to flip the minus sign on 10 to a plus sign; grouping as (6x² + 4x) − (15x + 10) gives 2x(3x + 2) − 5(3x + 2) = (3x + 2)(2x − 5). Check Soraya's grouping by distributing the minus sign: −(15x − 10) = −15x + 10, which does not match −15x − 10. Group correctly by factoring −5 out of −15x − 10: 6x² + 4x − 15x − 10 = 2x(3x + 2) − 5(3x + 2). Factor out the shared binomial (3x + 2) to get (3x + 2)(2x − 5).
- The shared side has length 2x + 3 feet; Plot B is longer along its other dimension by x − 3 feet. Factor Plot A's area: 6x² + 5x − 6 = 6x² + 9x − 4x − 6 = 3x(2x + 3) − 2(2x + 3) = (2x + 3)(3x − 2). Factor Plot B's area: 8x² + 2x − 15 = 8x² + 12x − 10x − 15 = 4x(2x + 3) − 5(2x + 3) = (2x + 3)(4x − 5). Both plots share the side of length 2x + 3 feet. Subtract Plot A's other dimension from Plot B's other dimension: (4x − 5) − (3x − 2) = x − 3 feet, which is positive for all x > 4.
Tier 4 Challenge
- k = 14 gives (3x − 1)(x + 5); k = 2 gives (3x + 5)(x − 1); k = −2 gives (3x − 5)(x + 1); k = −14 gives (3x + 1)(x − 5); the largest possible perimeter is 8x + 8 meters, achieved by both k = 14 and k = 2. Multiply 3 × (−5) = −15. The four integer factor pairs of −15 are 15 and −1 with sum 14, 5 and −3 with sum 2, −5 and 3 with sum −2, and −15 and 1 with sum −14. Factor for each k: k = 14 gives (3x − 1)(x + 5) with perimeter 2(4x + 4) = 8x + 8 meters; k = 2 gives (3x + 5)(x − 1) with perimeter 2(4x + 4) = 8x + 8 meters. Both negative values, k = −2 and k = −14, give a perimeter of 2(4x − 4) = 8x − 8 meters, so the largest possible perimeter is 8x + 8 meters.
- Both answers are completely right and equivalent because distributing the outside −1 in Ronan's answer into (2x − 3) produces −2x + 3, which is Freja's factor (3 − 2x); both expand to −6x² + 7x + 3. Multiply −1 into the first factor of Ronan's expression: −1(2x − 3) = −2x + 3 = 3 − 2x. Replace −(2x − 3) with (3 − 2x) to get Freja's expression (3 − 2x)(3x + 1). Expand (3 − 2x)(3x + 1) = 9x + 3 − 6x² − 2x = −6x² + 7x + 3, confirming both are valid.
- Both containers share the base dimension 2x + 3 feet; Container A's other base dimension is 6x − 5 feet and Container B's is 6x − 7 feet; Container A has the greater base perimeter by 4 feet. Factor Container A's volume: x(12x² + 8x − 15) = x(12x² + 18x − 10x − 15) = x(2x + 3)(6x − 5). Its base perimeter is 2(2x + 3 + 6x − 5) = 16x − 4 feet. Factor Container B's volume: x(12x² + 4x − 21) = x(12x² + 18x − 14x − 21) = x(2x + 3)(6x − 7). Its base perimeter is 2(2x + 3 + 6x − 7) = 16x − 8 feet. Subtract the base perimeters: (16x − 4) − (16x − 8) = 4 feet.
- Length is 4x + 5 meters and width is 2x + 3 meters, from the factored form (4x + 5)(2x + 3). Let u represent the binomial 2x + 1, so the area expression becomes 2u² + 7u + 6. Factor 2u² + 7u + 6: since 4 × 3 = 12 and 4 + 3 = 7, 2u² + 4u + 3u + 6 = (2u + 3)(u + 2). Substitute 2x + 1 back in for u and simplify: (2(2x + 1) + 3)((2x + 1) + 2) = (4x + 5)(2x + 3).
- Factored form is y = −2t(2t − 3)(4t − 1); the zeroes are t = 0, t = 1/4 and t = 3/2; the interval from 1/4 to 3/2 has length 5/4. Factor out the negative greatest common factor −2t: y = −2t(8t² − 14t + 3). For 8t² − 14t + 3, the numbers −12 and −2 multiply to 24 and add to −14: 8t² − 12t − 2t + 3 = 4t(2t − 3) − 1(2t − 3) = (2t − 3)(4t − 1), giving y = −2t(2t − 3)(4t − 1). Setting each factor equal to 0 gives t = 0, t = 1/4 and t = 3/2. At t = 1, y = −2(1)(−1)(3) = 6, so y is positive from t = 1/4 to t = 3/2, an interval of 3/2 − 1/4 = 5/4.
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