Factoring Trinomials Practice Problems

What this problem type covers

Each problem is a quadratic trinomial to be factored into binomials. The easier bands have a leading coefficient of 1 and small roots; the harder ones add a common factor to pull out first and larger numbers to search through.

Example problems

One example from each difficulty band, with worked solutions.

These problems are generated automatically. The answers are derived as each problem is built and are checked by tests, but mistakes are still possible — please tell us if you find one.

1. 1 · ★1
   Factor the expression.
   x² − 5x
   • ①(x − 1)(x − 5)
   • ②(x + 1)(x − 5)
   • ③x(x − 6)
   • ④x(x + 5)
   • ⑤x(x − 5)
   Show solution

   ⑤x(x − 5)

   1. Look for two numbers whose product is 0 and whose sum is −5.
   2. Those numbers are 0 and −5.
   3. So the factorisation is x(x − 5)
2. 2 · ★2
   Factor the expression.
   x² − 12x + 35
   • ①(x − 8)(x − 5)
   • ②(x − 7)(x − 5)
   • ③(x + 7)(x + 5)
   • ④(x + 7)(x − 5)
   • ⑤(x − 7)(x + 5)
   Show solution

   ②(x − 7)(x − 5)

   1. Look for two numbers whose product is 35 and whose sum is −12.
   2. Those numbers are −7 and −5.
   3. So the factorisation is (x − 7)(x − 5)
3. 3 · ★3
   Factor the expression.
   x² + 2x − 3
   • ①(x + 1)(x + 3)
   • ②(x + 1)(x − 3)
   • ③(x − 1)(x − 3)
   • ④(x − 2)(x + 3)
   • ⑤(x − 1)(x + 3)
   Show solution

   ⑤(x − 1)(x + 3)

   1. Look for two numbers whose product is −3 and whose sum is 2.
   2. Those numbers are −1 and 3.
   3. So the factorisation is (x − 1)(x + 3)
4. 4 · ★4
   Factor the expression.
   2x² − 14x − 16
   • ①2x(x − 8)
   • ②2(x + 1)(x + 8)
   • ③2(x − 1)(x + 8)
   • ④2(x + 1)(x − 8)
   • ⑤(x + 1)(x − 8)
   Show solution

   ④2(x + 1)(x − 8)

   1. Take out the common factor 2 first.
   2. Look for two numbers whose product is −8 and whose sum is −7.
   3. Those numbers are 1 and −8.
   4. So the factorisation is 2(x + 1)(x − 8)
5. 5 · ★5
   Factor the expression.
   2x² − 30x + 108
   • ①2(x − 9)(x − 6)
   • ②2(x − 9)(x + 6)
   • ③2(x + 9)(x + 6)
   • ④2(x − 10)(x − 6)
   • ⑤(x − 9)(x − 6)
   Show solution

   ①2(x − 9)(x − 6)

   1. Take out the common factor 2 first.
   2. Look for two numbers whose product is 54 and whose sum is −15.
   3. Those numbers are −9 and −6.
   4. So the factorisation is 2(x − 9)(x − 6)

Where people usually go wrong

• Getting the signs backwards. In (x − 3)(x + 5) the roots are 3 and −5, not −3 and 5 — the factor and the root always carry opposite signs.
• Forgetting the common factor. 2x² − 12x + 18 is 2(x − 3)², and an answer that skips the 2 is not equal to the original expression.
• Finding numbers with the right product but the wrong sum. Both conditions have to hold at once; checking only the product is the usual shortcut that fails.
• Stopping when one factor is still factorable. Factoring means factoring completely.
• Expanding to check but only multiplying the first and last terms. The middle term is exactly where sign errors hide.