If x² - px + 56 has factors differing by 1, then p equals?
- 13
- 14
- 15
- None of these
Explanation
Let the factors be (x - a) and (x - (a+1)) because they differ by 1.
Multiply them:
(x - a)(x - a - 1) = x^2 - (2a + 1)x + a(a + 1)
Compare with x^2 - px + 56:
a(a + 1) = 56
p = 2a + 1
Solve a(a + 1) = 56
a^2 + a - 56 = 0
(a + 8)(a - 7) = 0
So a = 7 or a = -8
If a = 7, then p = 2_7 + 1 = 15
If a = -8, then p = 2_(-8) + 1 = -15
Last verified on 23-05-2026
Related MCQs
- 2/x + 7/(x + 4)
- 1/x + 8/(x + 4)
- 8/x + 1/(x + 4)
- None of these
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