Mathematics

Example: Apply the division algorithm to find the quotient and remainder on dividing p(x) by g(x) as given : p(x) = x⁴ – 3x² + 4x + 5, g (x) = x² + 1 – x

Solution: We have,
p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

We stop here since degree of (8) < degree of (x² – x + 1). So, quotient = x² + x – 3, remainder = 8. Therefore,
 Quotient × Divisor + Remainder
= (x² + x – 3) (x² – x + 1) + 8
= x⁴ – x³ + x² + x³ – x² + x – 3x² + 3x – 3 + 8
= x⁴ – 3x² + 4x + 5 = Dividend
Therefore the Division Algorithm is verified.

Example: Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm. t² – 3; 2t⁴ + 3t³ – 2t² – 9t – 12.

Solution: We divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3

Here, remainder is 0, so t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.
2t⁴ + 3t³ – 2t² – 9t – 12 = (2t² + 3t + 4) (t² – 3)

Example: Obtain all the zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are  and –.

Solution: Since two zeroes are  and –, or we can say that the roots of equation are 
⇒ x =, x = –
⇒ = x² – 
⇒(3x² – 5) is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial and (3x² – 5). So,

⇒ 3x⁴ + 6x³ – 2x² – 10x – 5
= (3x² – 5) (x² + 2x + 1) + 0
Quotient = x² + 2x + 1 = (x + 1)²
Zeroes of (x + 1)² are –1, –1. Hence, all its zeroes are , – , –1, –1.

Example: On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (–2x + 4), respectively. Find g(x).

Solution: p(x) = x³ – 3x² + x + 2

q(x) = (x – 2) and r (x) = (–2x + 4)
By Division Algorithm, we know that p(x) = q(x) × g(x) + r(x). Therefore,
x³ – 3x² + x + 2 = (x – 2) × g(x) + (–2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4 = (x – 2) × g(x)
⇒ g(x) =
On dividing x³ – 3x² + 3x – 2 by (x – 2), we get g(x)

Hence, other polynomial is g(x) = x² – x + 1. which is a type of quadratic equation.