Mathematics

Example: Give examples of polynomials p(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x) , (ii) deg q(x) = deg r(x), (iii) deg q(x) = 0

Solution:
(i) Let q(x) = 3x² + 2x + 6, degree of q(x) = 2
p(x) = 12x² + 8x + 24, degree of p(x) = 2
Here, deg p(x) = deg q(x)

(ii) p(x) = x⁵ + 2x⁴ + 3x³ + 5x² + 2
q(x) = x² + x + 1, degree of q(x) = 2
g(x) = x³ + x² + x + 1
r(x) = 2x² – 2x + 1, degree of r(x) = 2
Here, deg q(x) = deg r(x)

(iii) Let p(x) = 2x⁴ + 8x³ + 6x² + 4x + 12
q(x) = 2, degree of q(x) = 0
g(x) = x⁴ + 4x³ + 3x² + 2x + 6
r(x) = 0
Here, deg q(x) = 0

Example: If the zeroes of polynomial x³ – 3x² + x + 1 are a – b, a , a + b. Find a and b.

Solution: Since, a – b, a, a + b are zeros of polynomial x³ – 3x² + x + 1. Also,
product (a – b) a(a + b) = –1
⇒ (a² – b²) a = –1 … (1) and 
sum of zeroes is (a – b) + a + (a + b) = 3
⇒ 3a = 3 ⇒ a = 1 …(2)
by the equation (1) and (2)
⇒ (1 – b²)1 = –1
⇒ 2 = b² ⇒ b = ±√2
Therefore, a = –1 & b = ±√2 

Example: If two zeroes of the polynomial x⁴ – 6x³ –26x² + 138x – 35 are 2 ± √3, Find other zeroes.

Solution: Since, 2 ± √3 are zeroes od a polynomial.
⇒ x = 2 ± √3
⇒ x – 2 = ±√3 (squaring both sides)
⇒ (x – 2)² = 3
⇒ x² + 4 – 4x = 3
⇒ x² + 4 – 4x – 3 = 0
⇒ x² – 4x + 1 = 0, is a factor of given polynomial. Now the other factor is given by dividing polynomial. i.e.   
=

Therefore, other factors = x² – 2x – 35
= x² – 7x + 5x – 35 = x(x – 7) + 5(x – 7)
= (x – 7) (x + 5)
⇒ other zeroes are (x – 7) = 0 ⇒ x = 7
x + 5 = 0 ⇒ x = – 5.

Example: If the polynomial x⁴ – 6x³ + 16x² –25x + 10 is divided by another polynomial x² –2x + k, the remainder comes out to be x + a, find the value of k and a.

Solution: 

According to questions, remainder is x + a, Therefore, coefficient of x = 1
⇒ 2k – 9 = 1
⇒ k = (10/2) = 5
Also constant term = a 
⇒ k² – 8k + 10 = a ⇒ (5)² – 8(5) + 10 = a
⇒ a = 25 – 40 + 10
⇒ a = – 5 
Therefore, the valur of k is 5 and a is –5.