# Polynomials-Principle Of Undetermined Coefficients

## Single Polynomial

If a polynomial of degree *n* in *x* vanishes for more than *n* different values *x* the coefficients of each power of *x* must be zero.
$$P(x) = c_0x^n + c_1x^{n-1} + c_2x^{n-2} + c_3x^{n-3} + .... c_{n-1}x + c_n ……(1) $$
If \(P(x) = 0\) when x equals each of the unequal values \(1, 2, 3, 4, ..... n\)

Then \((x-1),(x-2),(x-3), ........ (x-n)\) are all factors of \(P(x)\) so we can write
$$P(x) = N(x-1)(x-2)(x-3) ........ (x-n) ……(2)$$
where N is a numerical constant to be determined the same as in Worked Example 11 discussed earlier.

By expanding (2) then \(n^{th}\) term is \(c_0x^n\) so \(N = c_o\) and (2) becomes

$$P(x) = c_o(x-1)(x-2)(x-3), ........ (x-n) ……(3) $$
Let \(\beta\) another value of x that makes \(P(x)\) vanish, i.e. \((x - \beta) = 0\) , then
$$P(x) = c_o(\beta-1)(\beta-2)(\beta-3), ........ (\beta-n) ……(4) $$
Since none of the \((\beta-1)(\beta-2)(\beta-3), ........ (\beta-n)\) vanish then \(c_0 = 0\)

So (1) can be written
$$P(x) = c_1x^{n-1} + c_2x^{n-2} + c_3x^{n-3} + .... c_{n-1}x + c_n$$
By repeating the argument to show \(c_0 = 0\) it can be shown that \(c_1 = 0\), and again repeated to show all all remaining coefficients are also zero.

## Two Polynomials

If two polynomials of degree *n* in *x* are equal for more than *n* values of *x* they are equal for all values of *x*.

If
$$c_0x^n + c_1x^{n-1} + c_2x^{n-2} + c_3x^{n-3} + .... c_{n-1}x + c_n ……(5)$$
$$d_0x^n + d_1x^{n-1} + d_2x^{n-2} + d_3x^{n-3} + .... d_{n-1}x + d_n ……(6)$$
are equal for more than *n* values of *x*, then
$$(c_0 - d_0)x^n + (c_1 - d_1)x^{n-1} + (c_2 - d_2)x^{n-2} + (c_3 - d_3)x^{n-3} + .... (c_{n-1} - d_{n-1})x + (c_n - d_n)$$
vanishes for more than *n* values of *x* then as shown in previous section all coefficients must equal zero.
$$\therefore (c_0 - d_0)=0, (c_1 - d_1)=0, (c_2 - d_2)=0, (c_3 - d_3)=0, .... (c_{n-1} - d_{n-1})=0, (c_n - d_n)=0$$
$$c_0 = d_0, c_1 = d_1 = 0, c_2 = d_2, c_3 = d_3, .... c_{n-1} = d_{n-1}, c_n = d_n$$
\(\therefore\) equations (5) & (6) are identical and so equal for all values of x.

**So if two polynomials in x are equal for all values of x we may equate the the coefficients of the like powers of x.**

### Example 13.

Find the relation between *q* and *r* so that \(x^3 + 3px^2 + qx + r\) shall be a perfect cube for all values of *x*.

\((x + a)^3 = (x^3 + 3ax^2 + 3a^2x + a^3)\) is a perfect cube for all real values of *a*.

Let
$$x^3 + 3px^2 +qx + r = x^3 + 3ax^2 + 3a^2x + a^3$$
Equate coeff.s of *x* and terms independent of *x*
$$q = 3a^2\text{ and }r = a^3$$
Cube *q* and square *r*
$$q^3 = 27a^6 \text{ and } r^2 = a^6 $$
$$\therefore \frac{q^3}{r^2} = \frac{27a^6}{a^6}=27$$
$$\therefore q^3 = 27r^2$$