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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=1, Mode=simple

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

1.   x 2 + 3x + 2
2.   x 2 + 7x + 10

Complexity=3, Mode=prefix

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

1.   x 2 - 4x - 32
2.   - 5x 3 + 55x 2 + 130x

Complexity=4, Mode=frontConst

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

1.   - 6x 3 - 111x 2 + 246x
2.   - 2x 2 + 20x + 78

Complexity=5, Mode=includeY

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

1.   - 5x 3 + 32x 2y - 51xy 2
2.   5x 3 + 40x 2y + 35xy 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.

1.   6w 3x 12z 3 + 114w 2x 8yz 2 + 204wx 4y 2z
2.   - 18w 3x 9z 3 + 537w 2x 6yz 2 + 366wx 3y 2z

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.

#ProblemCorrect AnswerYour Answer
1x 2 + 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)
#ProblemCorrect AnswerYour Answer
2x 2 + 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.

#ProblemCorrect AnswerYour Answer
1x 2 - 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)
#ProblemCorrect AnswerYour Answer
2- 5x 3 + 55x 2 + 130x
Solution
Factor out - 5x to get
  - 5x(x 2 - 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.

#ProblemCorrect AnswerYour Answer
1- 6x 3 - 111x 2 + 246x
Solution
Factor out - 3x to get
  - 3x(2x 2 + 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)
#ProblemCorrect AnswerYour Answer
2- 2x 2 + 20x + 78
Solution
Factor out - 2 to get
  - 2(x 2 - 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.

#ProblemCorrect AnswerYour Answer
1- 5x 3 + 32x 2y - 51xy 2
Solution
Factor out -x to get
  -x(5x 2 - 32xy + 51y 2)

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)
#ProblemCorrect AnswerYour Answer
25x 3 + 40x 2y + 35xy 2
Solution
Factor out 5x to get
  5x(x 2 + 8xy + 7y 2)

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.

#ProblemCorrect AnswerYour Answer
16w 3x 12z 3 + 114w 2x 8yz 2 + 204wx 4y 2z
Solution
Factor out 6wx 4z to get
  6wx 4z(w 2x 8z 2 + 19wx 4yz + 34y 2)

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 19wx 4yz.
Test 1, 1, 1, 34: (wx 4z + y)(wx 4z + 34y)
  34wx 4yz + wx 4yz = 35wx 4yz No
Test 1, 2, 1, 17: (wx 4z + 2y)(wx 4z + 17y)
  17wx 4yz + 2wx 4yz = 19wx 4yz This matches!
Result: 6wx 4z(wx 4z + 2y)(wx 4z + 17y)
#ProblemCorrect AnswerYour Answer
2- 18w 3x 9z 3 + 537w 2x 6yz 2 + 366wx 3y 2z
Solution
Factor out - 3wx 3z to get
  - 3wx 3z(6w 2x 6z 2 - 179wx 3yz - 122y 2)

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 - 179wx 3yz.
Test 1, 1, 6, -122: (wx 3z + y)(6wx 3z - 122y)
  - 122wx 3yz + 6wx 3yz = - 116wx 3yz No
Test 1, 2, 6, -61: (wx 3z + 2y)(6wx 3z - 61y)
  - 61wx 3yz + 12wx 3yz = - 49wx 3yz No
Test 1, 61, 6, -2: (wx 3z + 61y)(6wx 3z - 2y)
  - 2wx 3yz + 366wx 3yz = 364wx 3yz No
Test 1, 122, 6, -1: (wx 3z + 122y)(6wx 3z - y)
  -wx 3yz + 732wx 3yz = 731wx 3yz No
Test 1, -1, 6, 122: (wx 3z - y)(6wx 3z + 122y)
  122wx 3yz + - 6wx 3yz = 116wx 3yz No
Test 1, -2, 6, 61: (wx 3z - 2y)(6wx 3z + 61y)
  61wx 3yz + - 12wx 3yz = 49wx 3yz No
Test 1, -61, 6, 2: (wx 3z - 61y)(6wx 3z + 2y)
  2wx 3yz + - 366wx 3yz = - 364wx 3yz No
Test 1, -122, 6, 1: (wx 3z - 122y)(6wx 3z + y)
  wx 3yz + - 732wx 3yz = - 731wx 3yz No
Test 2, 1, 3, -122: (2wx 3z + y)(3wx 3z - 122y)
  - 244wx 3yz + 3wx 3yz = - 241wx 3yz No
Test 2, 2, 3, -61: (2wx 3z + 2y)(3wx 3z - 61y)
  - 122wx 3yz + 6wx 3yz = - 116wx 3yz No
Test 2, 61, 3, -2: (2wx 3z + 61y)(3wx 3z - 2y)
  - 4wx 3yz + 183wx 3yz = 179wx 3yz No
Test 2, 122, 3, -1: (2wx 3z + 122y)(3wx 3z - y)
  - 2wx 3yz + 366wx 3yz = 364wx 3yz No
Test 2, -1, 3, 122: (2wx 3z - y)(3wx 3z + 122y)
  244wx 3yz + - 3wx 3yz = 241wx 3yz No
Test 2, -2, 3, 61: (2wx 3z - 2y)(3wx 3z + 61y)
  122wx 3yz + - 6wx 3yz = 116wx 3yz No
Test 2, -61, 3, 2: (2wx 3z - 61y)(3wx 3z + 2y)
  4wx 3yz + - 183wx 3yz = - 179wx 3yz This matches!
Result: - 3wx 3z(2wx 3z - 61y)(3wx 3z + 2y)
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