Level 2 · Intermediate ← Back to lesson

Factoring by Grouping: Practice

12 multiple-choice questions, progressively harder.

Level 2 · Intermediate 0 / 12 answered
Question 1 of 12
  1. 1

    Factor x3−3x2−2x+6x^3 - 3x^2 - 2x + 6 by grouping.

    Answer choices for question 1
  2. 2

    Factor x3+2x2−5x−10x^3 + 2x^2 - 5x - 10 by grouping.

    Answer choices for question 2
  3. 3

    Factor 2x2+7x+62x^2 + 7x + 6 by the AC method.

    Answer choices for question 3
  4. 4

    Factor x3−5x2+3x−15x^3 - 5x^2 + 3x - 15 by grouping.

    Answer choices for question 4
  5. 5

    The polynomial x3+4x2−3x−12x^3 + 4x^2 - 3x - 12 factors by grouping. What binomial is common to both pairs?

    Answer choices for question 5
  6. 6

    For 3x2−5x−23x^2 - 5x - 2, the product is ac=−6ac = -6 and the sum is −5-5. Which numbers split the middle term?

    Answer choices for question 6
  7. 7

    Reorder and factor 2x3−5x+6x2−152x^3 - 5x + 6x^2 - 15.

    Answer choices for question 7
  8. 8

    Factor 4x2+8x+3x+64x^2 + 8x + 3x + 6 (already split) by grouping.

    Answer choices for question 8
  9. 9

    Factor 2x2−5x−32x^2 - 5x - 3 by the AC method.

    Answer choices for question 9
  10. 10

    The second pair of x3−4x2−5x+20x^3 - 4x^2 - 5x + 20 is −5x+20-5x + 20. Factoring it so the binomial matches x−4x - 4 gives:

    Answer choices for question 10
  11. 11

    Factor 2x2−7x+62x^2 - 7x + 6 by the AC method.

    Answer choices for question 11
  12. 12

    Factor 4x2−4x−34x^2 - 4x - 3 by the AC method.

    Answer choices for question 12