Algebra IOQM theory - 2
1. If f(x), g(x) are polynomials with degree m,n each:
A. Degree of f(x) + g(x) = max(m,n)
B. Degree of f(x).g(x) = m + n
C. Degree of f(g(x)) = m.n
2. Leading coefficient = coefficient of degree defining term
3. Monic polynomial => leading coefficient = 1
4. Dividend = Divisor * Quotient + Remainder
f(x) = h(x).Q(x) + R(x)
where h(x) is divisor of f(x) leaving quotient as Q(x) and remainder as R(x).
f(x) has degree m.
h(x) has degree n.
then:
degree of Q(x) = m - n.
degree of R(x) < n
5. Factor theorem:
A polynomial f(x) has factor (x-a) if f(a) = 0.
f(x) = (x-a).Q(x) + r
If r = 0 then (x-a) is a factor of f(x).
f(a) = 0.Q(a) + 0
f(a) = 0
6. Remainder theorem:
Remainder when f(x) is divided by (x-a) is f(a).
Proof:
f(x) = (x-a).Q(x) + R(x)
But R(x) needs to have degree 0 since divisor is of degree 1.
So R(x) = r = constant.
f(a) = (a-a).Q(a) + r = 0 + r = r
=> f(a) = r. Hence proved.
Ex1:
f(x) = 3x^2 - 2x + 1
Divide f(x) by x-2 and find the remainder.
Solution:
By remainder theorem f(2) is the remainder when f(x) is divided by (x-2).
f(2) = 9
Ex2:
Find remainder when x^2008 + 2008.x + 2008 is divided by (x+1).
Solution:
f(-1) = 1 - 2008 + 2008 = 1
Ex3:
Find remainder when x^100 + x^99 + x^98 + ... x^10 is divided by:
a) x-1
b) x+1
Solution:
a) f(1) = 91
b) f(-1) = 1
Ex4:
f(x) is a polynomial of degree 5.
Remainder when divided by (x-1) is 5.
Remainder when divided by (x-2) is 3.
Find remainder when divided by (x-1).(x-2)
Solution:
f(1) = 5
f(2) = 3
f(x) = (x-1)(x-2).Q(x) + R(x)
R(x) will be of degree 1 since divisor is of degree 2.
So
f(x) = (x-1)(x-2).Q(x) + ax + b
f(1) = 5 = a + b
f(2) = 3 = 2a + b
=> a = -2, b = 7
R(x) = -2x + 7
Ex5:
f(x) is a polynomial.
Remainder when divided by (x+1) is 2.
Remainder when divided by (x-2) is 7.
Remainder when divided by (x) is 1.
Find remainder when divided by x.(x+1).(x-2)
Solution:
f(-1) = 2
f(2) = 7
f(0) = 1
f(x) = x.(x+1).(x-2).Q(x) + R(x)
R(x) will be quadratic now: ax^2 + bx + c
2 = a - b + c
7 = 4a + 2b + c
1 = c
=>
a-b = 1
2a+b = 3
3a = 4 => a = 4/3, b=1/3
R(x) = 4.x^2/3 + x/3 + 1
Ex6:
f(x) is monic polynomial of degree 3.
f(1) = 1, f(2) = 2, f(3) = 3
Then f(4) = ?
Solution:
Let's create g(x) = f(x) - x.
Now g(x) is also a monic polynomial of degree 3.
g(1) = g(2) = g(3) = 0
=> g(x) = (x-1)(x-2)(x-3)
g(4) = 6
f(4) = g(4) + 4 = 10
Ex6:
f(x) is polynomial of degree 4 with leading coefficient = 2.
f(0) = -1, f(1) = 0, f(2) = 3, f(3) = 8
Then f(-1) = ?
Solution:
g(x) = f(x) - (x^2 - 1)
g(x) = 2.x.(x-1).(x-2).(x-3)
g(-1) = 48
f(-1) = g(-1) + (1-1) = 48
Ex7:
Find all polynomials such that
x.P(x-1) = (x-4).P(x)
for every x.
Solution:
x = 0 => P(0) = 0
x = 1 => P(1) = P(0) = 0
Similarly P(2) = P(3) = 0
But we can't say this for x = 4.
Why? Because:
4.P(4-1) = 0.P(4)
0 = 0.P(4)
Here both sides are 0. So P(4) can't be deduced.
So a general form for P(x) is:
P(x) = K.x.(x-1).(x-2).(x-3) where K is a constant.
In that case, rather than multiplying by a constant, we should be multiplying by another polynomial?
Let's see why that's not the case.
Let P(x) = a.x^n + b.x^(n-1)...
In the original equation:
LHS = x.P(x-1) = x.(a.(x-1)^n + b.(x-1)^(n-1)...)
RHS = (x-4).P(x) = (x-4).(a.x^n + b.x^(n-1)...)
Coefficient of x^n should be same in both the sides.
In RHS, it is: -4a + b
In LHS, it is: b + a.(nC1).(-1) = b - an
b - an = b - 4a
=> n = 4
So it is indeed a 4-degree polynomial. And we found all the roots.
Ex8:
f(x) is a monic biquadratic such that f(1) = 10, f(2) = 20, f(3) = 30.
Find (f(12) + f(-8))/10.
Solution:
g(x) = f(x) - 10x
g(x) has 1,2,3 as roots.
g(x) = (x-1).(x-2).(x-3).(x-a)
f(x) = g(x) + 10x
f(12) = 120 + 9.10.11.(12-a)
f(-8) = -9*-10*-11(-8-a)-80
Now terms with 'a' will cancel out.
f(12) + f(-8) = 40 + 9.10.11(12 + 8) = 40 + 19800 = 19840
Answer: 1984
Ex9:
f(x) = 2.x^8 + 14.x^2 + 2.x^3 + 1997
Find number of integer zeroes for f(x).
Solution:
None.
As even terms won't be able to cancel out the odd term.
Vieta's formulas.
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