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Polynomial Equation of Degree n A polynomial equation of degree n is an important topic of the IIT Mathematics syllabus. It comes under the head of Quadratic equations. This is a simple topic which can fetch you direct questions which are too scoring. Let us first discuss what exactly we mean by a polynomial equation of degree n: Consider the equation a_{n}x^{n} + a_{n–1}x^{n–1} + a_{n–2}x^{n–2} +……+ a_{1}x + a_{0} = 0 … (1) (a_{0}, a_{1}, … , a_{n} are real coefficients and a_{n} ≠ 0). Let a_{1}, a_{2}, ……, a_{n} be the roots of equation (1). Then by simplifying the obtained equations and hence, comparing the coefficients of like powers of x, we get a_{1} + a_{2} + a_{3} +……+ a_{n} = -a_{n-1}/ a_{n} a_{1}a_{2} + a_{1}a_{3} + a_{1}a_{4} + …… + a_{2}a_{3} +……+ a_{n–1}a^{n} = a_{n-2}/ a_{n}, …………………………………… a_{1}a_{2} …… a_{r} + … + a_{n–r+1}a_{n–r+2 }… a_{n} = (–1)^{r} a_{n-r}/a_{n}, …………………………………… a_{1}a_{2} …… a_{n} = (–1)^{n} a_{0}/a_{n}. On the similar lines let us consider an equation ax^{4} + bx^{3} + cx^{2} + dx + e = 0 Let us assume the roots of the equation as a, b, g and d. Then as derived above, we obtain the following equations: a + b + g + d = –b/a, ab + ag + ad + bg + bd + gd = c/a, abg + abd + agd + bgd = –d/a, abgd = e/a. These results are in fact the formulae to be applied while dealing with the numerical based on these. Some Key points:
A polynomial equation of degree n is an important topic of the IIT Mathematics syllabus. It comes under the head of Quadratic equations. This is a simple topic which can fetch you direct questions which are too scoring.
Let us first discuss what exactly we mean by a polynomial equation of degree n:
Consider the equation
a_{n}x^{n} + a_{n–1}x^{n–1} + a_{n–2}x^{n–2} +……+ a_{1}x + a_{0} = 0 … (1)
(a_{0}, a_{1}, … , a_{n} are real coefficients and a_{n} ≠ 0).
Let a_{1}, a_{2}, ……, a_{n} be the roots of equation (1).
Then by simplifying the obtained equations and hence, comparing the coefficients of like powers of x, we get
a_{1} + a_{2} + a_{3} +……+ a_{n} = -a_{n-1}/ a_{n}
a_{1}a_{2} + a_{1}a_{3} + a_{1}a_{4} + …… + a_{2}a_{3} +……+ a_{n–1}a^{n} = a_{n-2}/ a_{n},
……………………………………
a_{1}a_{2} …… a_{r} + … + a_{n–r+1}a_{n–r+2 }… a_{n} = (–1)^{r} a_{n-r}/a_{n},
a_{1}a_{2} …… a_{n} = (–1)^{n} a_{0}/a_{n}.
On the similar lines let us consider an equation ax^{4} + bx^{3} + cx^{2} + dx + e = 0
Let us assume the roots of the equation as a, b, g and d.
Then as derived above, we obtain the following equations:
a + b + g + d = –b/a,
ab + ag + ad + bg + bd + gd = c/a,
abg + abd + agd + bgd = –d/a,
abgd = e/a.
These results are in fact the formulae to be applied while dealing with the numerical based on these.
A polynomial equation of degree n has n roots (real or imaginary).
If all the coefficients are real then the imaginary roots occur in pairs i.e. number of complex roots is always even.
If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at least one of the roots will be real.
If a is repeated root (repeating r times) of a polynomial equation f(x) = 0 of degree n i.e. f(x) = (x – a)^{r} g(x), where g(x) is a polynomial of degree n – r and g(a) ≠ 0, then f(a) = f’(a) = f’’(a) = … = f^{(r–1)}(a) = 0 and f^{r}(a) ≠ 0.
If a polynomial equation of degree n has n + 1 roots say x_{1}, …, x_{n+1}, where x_{i }≠ x_{j} if i ≠ j, then the polynomial is identically zero i.e. p(x) = 0, ∀ x ∈ R. this can also be stated as the coefficients a_{0}, ……, a_{n} are all zero.
If p(a) and p(b) (a < b) are of opposite signs, then p(x) = 0 has odd number of roots in (a, b) i.e. it will have at least one root in (a, b).
If coefficients in p(x) have ‘m’ changes in signs, then p(x) = 0 can have at most ‘m’ positive real roots and if p(–x) has ‘t’ changes in sign, then p(x) = 0 can have at most ‘t’ negative real roots. By this we can find maximum number of real roots and minimum number of complex roots of a polynomial equations p(x) = 0.
The remainder theorem as the name suggests, talks about the remainder when a polynomial is divided by some linear factor. This division obviously results in some quotient as well as remainder. The remainder theorem can be stated as:
When a polynomial say ‘p(x)’ is divided by some linear factor say (x-a), we obtain a remainder of the form p(a). In particular, if p(a) = 0, then (x-a) is a factor of p(x).
The remainder theorem is extremely useful in finding the roots of a polynomial. It helps in identification of a factor which divides the polynomial completely. In large degree polynomials, it would have been a tedious task to find the roots without the remainder theorem. Now, with this theorem we can find out a single root which can then be used synthetically to obtain a smaller polynomial for which the process is repeated until all the roots are obtained.
Let us consider an example based on the remainder theorem:
Illustration: Find all the roots of the equation x^{4}+7x^{3}+3x^{2}-63x-108.
Solution: Seeing the polynomial it is clear that the polynomial will have 5 roots.
We aim to find those values of x for which P(x) = 0.
We first start by substituting x =0, but P (0) = 108 which is not zero and so 0 is not a root of the polynomial. Now we proceed further and try to substitute x = -3. This equals zero and hence, it proves that -3 is a root of the given polynomial. So, by the remainder theorem (x + 3) is a factor of the polynomial. So now we can use synthetic division to get a smaller polynomial
We obtain a new polynomial; let’s call this polynomial f(x)
f(x) = x^{3}+4x^{2}-9x-36.
Now, this is the new reduced polynomial and again we proceed by trial and error. This gives x = 3 as the root and so (x-3) is a factor.
Next we divide f(x) by (x - 3)
We obtain a new polynomial g(x) and a remainder of zero
f(x) = 4x^{2}+7x+12
Repeating the same process again we get two more roots -4 and -3. Hence, the polynomial P(x) can be written as P(x) = (x+3)(x+3)(x-4)(x-3).
Hence, the roots of the polynomial are x = {4, -3, -3, 3}.
Illustration: Let a and b be two roots of the equation x^{3} + px^{2} + qx + r = 0 satisfying the relation ab + 1 = 0. Prove that r^{2} + pr + q + 1 = 0. (r ≠ 0)
Solution: Given equation is x^{3} + px^{2} + qx + r = 0.
Let the third root be c, so that abc = –r. …… (1)
Given ab = –1
from (1) c = r, which is a root of the given equation.
(r)^{3} + p(r)^{2} + q(r) + r = 0 Þ r^{2} + pr + q + 1 = 0.
Illustration: If b^{2} < 2ac, then prove that ax^{3} + bx^{2} + cx + d = 0 has exactly one real root.
Solution: Let a, b, g be the roots of ax^{3} + bx^{2} + cx + d = 0.
Then a + b + g = b/a,
ab + bg + ga = c/a,
abg = -d/a
This gives a^{2} + b^{2} + g^{2} = (a + b + g)^{2} –2(ab + bg + ga) = b^{2}/a^{2} –2c/a
= (b^{2} – 2ac)/ a^{2}
This gives a^{2} + b^{2} + g^{2} < 0, which is not possible if all a, b, g are real. So at least one root is non-real, but complex roots occur in pairs. Hence given cubic equation has two non-real and one real roots.
Alternative Solution:
Let f(x) = ax^{3} + bx^{2} + cx + d and f(x) = 0 have all the roots real
So, f’(x) = 3ax^{2} + 2bx + c = 0 has two real roots.
But its discriminant
= (2b)^{2} – 4.3.ac = (b^{2} – 2ac) – ac < 0 (as b^{2} < 2ac)
which is a contradiction and so f(x) = 0 will not have all the roots real.
Related resources:
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