Math Practice Problems - Trinomial Factoring

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Trinomial Factoring - Sample Math Practice Problems

The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Answers to these sample questions appear at the bottom of the page. This page does not grade your responses.

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Complexity=3, Mode=prefix

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

Complexity=4, Mode=frontConst

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

Complexity=5, Mode=includeY

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

Complexity=6, Mode=hardest

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

Answers

Complexity=1, Mode=simple

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

#	Problem	Correct Answer	Your Answer
1	x² + 3x + 2
Solution The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is 2. Factors are: 1, 2 Now test combinations of these factors to get the middle coefficient of 3. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term 3x. Test 1, 1, 1, 2: (x + 1)(x + 2) 2x + x = 3x This matches! Result: (x + 1)(x + 2)

#	Problem	Correct Answer	Your Answer
2	x² + 7x + 10
Solution The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is 10. Factors are: 1, 2, 5, 10 Now test combinations of these factors to get the middle coefficient of 7. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term 7x. Test 1, 1, 1, 10: (x + 1)(x + 10) 10x + x = 11x No Test 1, 2, 1, 5: (x + 2)(x + 5) 5x + 2x = 7x This matches! Result: (x + 2)(x + 5)

Complexity=3, Mode=prefix

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

#	Problem	Correct Answer	Your Answer
1	x² - 4x - 32
Solution The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is -32. Factors are: ±1, 2, 4, 8, 16, 32 Now test combinations of these factors to get the middle coefficient of -4. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term ^-4x. Test 1, 1, 1, -32: (x + 1)(x - 32) ^-32x + x = ^-31x No Test 1, 2, 1, -16: (x + 2)(x - 16) ^-16x + 2x = ^-14x No Test 1, 4, 1, -8: (x + 4)(x - 8) ^-8x + 4x = ^-4x This matches! Result: (x + 4)(x - 8)

#	Problem	Correct Answer	Your Answer
2	^-5x³ + 55x² + 130x
Solution Factor out ^-5x to get ^-5x(x² - 11x - 26) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is -26. Factors are: ±1, 2, 13, 26 Now test combinations of these factors to get the middle coefficient of -11. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term ^-11x. Test 1, 1, 1, -26: (x + 1)(x - 26) ^-26x + x = ^-25x No Test 1, 2, 1, -13: (x + 2)(x - 13) ^-13x + 2x = ^-11x This matches! Result: ^-5x(x + 2)(x - 13)

Complexity=4, Mode=frontConst

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

#	Problem	Correct Answer	Your Answer
1	^-6x³ - 111x² + 246x
Solution Factor out ^-3x to get ^-3x(2x² + 37x - 82) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 2. Factors are: 1, 2 Back 'coefficient' is -82. Factors are: ±1, 2, 41, 82 Now test combinations of these factors to get the middle coefficient of 37. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term 37x. Test 1, 1, 2, -82: (x + 1)(2x - 82) ^-82x + 2x = ^-80x No Test 1, 2, 2, -41: (x + 2)(2x - 41) ^-41x + 4x = ^-37x No Test 1, 41, 2, -2: (x + 41)(2x - 2) ^-2x + 82x = 80x No Test 1, 82, 2, -1: (x + 82)(2x - 1) -x + 164x = 163x No Test 1, -1, 2, 82: (x - 1)(2x + 82) 82x + ^-2x = 80x No Test 1, -2, 2, 41: (x - 2)(2x + 41) 41x + ^-4x = 37x This matches! Result: ^-3x(x - 2)(2x + 41)

#	Problem	Correct Answer	Your Answer
2	^-2x² + 20x + 78
Solution Factor out ^-2 to get ^-2(x² - 10x - 39) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is -39. Factors are: ±1, 3, 13, 39 Now test combinations of these factors to get the middle coefficient of -10. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term ^-10x. Test 1, 1, 1, -39: (x + 1)(x - 39) ^-39x + x = ^-38x No Test 1, 3, 1, -13: (x + 3)(x - 13) ^-13x + 3x = ^-10x This matches! Result: ^-2(x + 3)(x - 13)

Complexity=5, Mode=includeY

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

#	Problem	Correct Answer	Your Answer
1	^-5x³ + 32x²y - 51xy²
Solution Factor out -x to get -x(5x² - 32xy + 51y²) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 5. Factors are: 1, 5 Back 'coefficient' is 51. Factors are: 1, 3, 17, 51 Now test combinations of these factors to get the middle coefficient of -32. Since it is negative, we want the negative versions of 51's factors. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term ^-32xy. Test 1, -1, 5, -51: (x - y)(5x - 51y) ^-51xy + ^-5xy = ^-56xy No Test 1, -3, 5, -17: (x - 3y)(5x - 17y) ^-17xy + ^-15xy = ^-32xy This matches! Result: -x(x - 3y)(5x - 17y)

#	Problem	Correct Answer	Your Answer
2	5x³ + 40x²y + 35xy²
Solution Factor out 5x to get 5x(x² + 8xy + 7y²) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is 7. Factors are: 1, 7 Now test combinations of these factors to get the middle coefficient of 8. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term 8xy. Test 1, 1, 1, 7: (x + y)(x + 7y) 7xy + xy = 8xy This matches! Result: 5x(x + y)(x + 7y)

Complexity=6, Mode=hardest

Factor the expression as completely as possible. For squared expressions, such as (x+1)², type as (x+1)^2.

#	Problem	Correct Answer	Your Answer
1	6w³x¹²z³ + 114w²x⁸yz² + 204wx⁴y²z
Solution Factor out 6wx⁴z to get 6wx⁴z(w²x⁸z² + 19wx⁴yz + 34y²) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 1. Factors are: 1 Back 'coefficient' is 34. Factors are: 1, 2, 17, 34 Now test combinations of these factors to get the middle coefficient of 19. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term 19wx⁴yz. Test 1, 1, 1, 34: (wx⁴z + y)(wx⁴z + 34y) 34wx⁴yz + wx⁴yz = 35wx⁴yz No Test 1, 2, 1, 17: (wx⁴z + 2y)(wx⁴z + 17y) 17wx⁴yz + 2wx⁴yz = 19wx⁴yz This matches! Result: 6wx⁴z(wx⁴z + 2y)(wx⁴z + 17y)

#	Problem	Correct Answer	Your Answer
2	^-18w³x⁹z³ + 537w²x⁶yz² + 366wx³y²z
Solution Factor out ^-3wx³z to get ^-3wx³z(6w²x⁶z² - 179wx³yz - 122y²) The factored form of the trinomial will take the form of (x + ?)(x + ?) Front coefficient is 6. Factors are: 1, 2, 3, 6 Back 'coefficient' is -122. Factors are: ±1, 2, 61, 122 Now test combinations of these factors to get the middle coefficient of -179. For each test, do a partial FOIL of (x + ?)(x + ?). Add the outer and inner products to see if that is equal to the middle term ^-179wx³yz. Test 1, 1, 6, -122: (wx³z + y)(6wx³z - 122y) ^-122wx³yz + 6wx³yz = ^-116wx³yz No Test 1, 2, 6, -61: (wx³z + 2y)(6wx³z - 61y) ^-61wx³yz + 12wx³yz = ^-49wx³yz No Test 1, 61, 6, -2: (wx³z + 61y)(6wx³z - 2y) ^-2wx³yz + 366wx³yz = 364wx³yz No Test 1, 122, 6, -1: (wx³z + 122y)(6wx³z - y) -wx³yz + 732wx³yz = 731wx³yz No Test 1, -1, 6, 122: (wx³z - y)(6wx³z + 122y) 122wx³yz + ^-6wx³yz = 116wx³yz No Test 1, -2, 6, 61: (wx³z - 2y)(6wx³z + 61y) 61wx³yz + ^-12wx³yz = 49wx³yz No Test 1, -61, 6, 2: (wx³z - 61y)(6wx³z + 2y) 2wx³yz + ^-366wx³yz = ^-364wx³yz No Test 1, -122, 6, 1: (wx³z - 122y)(6wx³z + y) wx³yz + ^-732wx³yz = ^-731wx³yz No Test 2, 1, 3, -122: (2wx³z + y)(3wx³z - 122y) ^-244wx³yz + 3wx³yz = ^-241wx³yz No Test 2, 2, 3, -61: (2wx³z + 2y)(3wx³z - 61y) ^-122wx³yz + 6wx³yz = ^-116wx³yz No Test 2, 61, 3, -2: (2wx³z + 61y)(3wx³z - 2y) ^-4wx³yz + 183wx³yz = 179wx³yz No Test 2, 122, 3, -1: (2wx³z + 122y)(3wx³z - y) ^-2wx³yz + 366wx³yz = 364wx³yz No Test 2, -1, 3, 122: (2wx³z - y)(3wx³z + 122y) 244wx³yz + ^-3wx³yz = 241wx³yz No Test 2, -2, 3, 61: (2wx³z - 2y)(3wx³z + 61y) 122wx³yz + ^-6wx³yz = 116wx³yz No Test 2, -61, 3, 2: (2wx³z - 61y)(3wx³z + 2y) 4wx³yz + ^-183wx³yz = ^-179wx³yz This matches! Result: ^-3wx³z(2wx³z - 61y)(3wx³z + 2y)

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