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Factorisation · 3 marks

If both x − 2 and x − 12 are factors of px2 + 5x + r, show that p = r.

Answer: Putting x = 2 and x = 12 gives 4p + r + 10 = 0 and p + 4r + 10 = 0. Subtracting, 3p − 3r = 0, so p = r (in fact p = r = −2).

Step-by-step solution

Given: x − 2 is a factor of px2 + 5x + r; x − 12 is a factor of px2 + 5x + r
To find: Show that p = r

Idea: A challenge question. If x − 2 is a factor, the expression is (x − 2) × (something). At x = 2 that factor becomes 0, so the whole expression becomes 0. The same is true for x − 12 at x = 12. This gives two equations in p and r.

  1. Let f(x) = px2 + 5x + r. Since x − 2 is a factor, f(x) = (x − 2) × (another factor). Putting x = 2 makes x − 2 = 0, so f(2) = 0. In the same way f(12) = 0.½ mark
  2. f(2) = 0:
    p(2)2 + 5(2) + r = 0
    4p + 10 + r = 0   …(1)½ mark
  3. f(12) = 0:
    p(12)2 + 5(12) + r = 0
    p4 + 52 + r = 0
    Multiply by 4: p + 10 + 4r = 0   …(2)1 mark
  4. Subtract (2) from (1):
    (4p + 10 + r) − (p + 10 + 4r) = 0
    3p − 3r = 0
    So p = r. Hence proved.1 mark
  5. Extra (not asked): putting r = p in (1) gives 5p + 10 = 0, so p = r = −2. The expression is −2x2 + 5x − 2.
From f(2) = 0 and f(12) = 0 we get 4p + r + 10 = 0 and p + 4r + 10 = 0, which give 3p = 3r, so p = r (both equal −2).

Check: With p = r = −2: −2x2 + 5x − 2 = −(2x2 − 5x + 2) = −(x − 2)(2x − 1) = −2(x − 2)(x − 12). Both x − 2 and x − 12 are factors ✓. Another way: the expression must be k(x − 2)(x − 12) = k(x2 − 52x + 1); then p = k and r = k, so p = r.

Answer to write in the exam

Let f(x) = px2 + 5x + r

f(2) = 0 [x − 2 is a factor]

p(2)2 + 5(2) + r = 0 ⇒ 4p + r + 10 = 0 …(1)

f(12) = 0 [x − 12 is a factor]

p(12)2 + 5(12) + r = 0 ⇒ p4 + 52 + r = 0 ⇒ p + 4r + 10 = 0 …(2)

(1) − (2): 3p − 3r = 0

∴ p = r Hence proved.

Common mistakes that cost marks

  • Substituting x = −2 and x = −12. For the factor x − 2 you put x = 2 (the value that makes the factor zero).
  • Squaring wrongly: writing p(12)2 = p2. It is p4.
  • Stopping at two equations without showing p = r clearly. Subtract the equations and write the conclusion “p = r” as the last line.

How this can come in the exam

MCQ (1 mark)

If x − 3 is a factor of x2 − 5x + k, the value of k is

  1. −6
  2. 6
  3. 3
  4. 15
Show answer

(B) 6
Put x = 3: 9 − 15 + k = 0, so k = 6. (Indeed x2 − 5x + 6 = (x − 2)(x − 3).)

Assertion–Reason (1 mark)

Assertion (A): If x + 1 is a factor of ax2 + bx + c, then a − b + c = 0.
Reason (R): If x − m is a factor of an expression in x, then the expression becomes 0 when x = m.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
x + 1 = x − (−1), so by R the expression is 0 at x = −1: a(−1)2 + b(−1) + c = a − b + c = 0. R explains A.

Try one yourself

If both x − 3 and x − 13 are factors of ax2 − 10x + c, show that a = c.

Show answer

x = 3: 9a − 30 + c = 0. x = 13: a9 − 103 + c = 0, i.e. a − 30 + 9c = 0. Subtracting: 8a − 8c = 0, so a = c (both equal 3; 3x2 − 10x + 3 = (x − 3)(3x − 1)).

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