of 5

# Polynomials - Polynomial Remainder Theorem

Given a cubic equation involving a single variable, we need to solve for the value of that variable, using the remainder theorem
Published on: Mar 4, 2016
Published in: Education

#### Transcripts - Polynomials - Polynomial Remainder Theorem

• 1. Polynomials Q1
• 2. Qn: Polynomial Remainder Theorem (a) (2, ) (b) (1, 2)  (2, ) (c) (- , 1)  (2, ) (d) (- , 1) Solve the inequality x3 – 5x2 + 8x – 4 > 0.
• 3. Soln: Polynomial Remainder Theorem Let a, b, c be the roots of this cubic equation a + b + c = 5 ab + bc + ca = 8 abc = 4 This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations.} The other approach is to use polynomial remainder theorem. Solve the inequality x3 – 5x2 + 8x – 4 > 0.
• 4. Soln: Polynomial Remainder Theorem If you notice, sum of the coefficients = 0  P(1) = 0  (x–1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x–1) and then factorising the resulting quadratic equation. (x – 1) (x – 2) (x – 2) > 0 Let us call the product (x – 1)(x – 2)(x – 2) a black box. If x is less than 1, the black box is a –ve number. Solve the inequality x3 – 5x2 + 8x – 4 > 0.
• 5. Soln: Polynomial Remainder Theorem If x is between 1 and 2, the black box is a +ve number. If x is greater than 2, the black box is a +ve number. Since we are searching for the regions where black box is a +ve number, the solution is as follows: 1 < x < 2 OR x > 2 Answer choice (b) Solve the inequality x3 – 5x2 + 8x – 4 > 0.