Finding p(0), p(1), p(−1), p(10) ബഹുപദത്തിൽ സംഖ്യകൾ
Writing p(3) means “put 3 in place of x in p(x) and work it out”. This page shows how to do that without slips – especially with negative numbers – and reveals four shortcuts: p(0), p(1), p(−1) and p(10) can be read almost straight from the coefficients.
- Powers: 2³ = 8, (−1)² = 1, (−1)³ = −1, 10³ = 1000.
- Coefficients (ഗുണകങ്ങൾ) are the fixed numbers multiplying the powers; the constant term has no x.
1. Substituting, step by step
Order of work: (1) put the number in brackets in place of x, (2) work out the powers, (3) multiply by the coefficients, (4) add.
- 3x² means 3 × (x²), not (3x)². With x = 2: 3 × 4 = 12, not 6² = 36.
- Always put a negative number in brackets: 3(−2)² = 3 × 4 = 12. Writing 3 × −2² invites the mistake −12.
2. Your own polynomial evaluator
Example: “4, 2, 3, 7” means 4x³ + 2x² + 3x + 7; “5, −8” means 5x − 8; “2, 0, −1” means 2x² − 1 (write 0 for a missing power).
3. p(0) – the constant term
Every term with an x becomes 0 when x = 0, so p(0) = the constant term. For ax³ + bx² + cx + d, p(0) = d.
4. p(1) – the sum of the coefficients
Every power of 1 is 1, so each term is just its coefficient: p(1) = sum of all the coefficients. For ax³ + bx² + cx + d, p(1) = a + b + c + d.
5. p(−1) – alternating signs
Each extra factor (−1) flips the sign: even powers give 1, odd powers give −1.
At x = −1, the coefficients of even powers (and the constant) keep their sign, those of odd powers change sign: for ax³ + bx² + cx + d, p(−1) = −a + b − c + d.
6. p(10) – place value in disguise
Our numbers are written in powers of 10: 4237 = 4 × 1000 + 2 × 100 + 3 × 10 + 7 = 4 × 10³ + 2 × 10² + 3 × 10 + 7. That is exactly p(10) for p(x) = 4x³ + 2x² + 3x + 7!
When all coefficients are digits 0–9, p(10) simply writes them side by side. (With a coefficient like 12 there is a “carry”, just as in addition.)
For ax³ + bx² + cx + d with digits a, b, c, d: p(10) is the number “abcd”. Example: 3x² + 6x + 1 → p(10) = 361.
Computers use the same idea with 2 in place of 10: the binary number 1011 is p(2) for p(x) = x³ + 0x² + x + 1, that is 8 + 2 + 1 = 11.