Question

B. Factor each polynomial completely. 1) p(x)= x3 + 2x2 – x - 2 2) p(x) = x3 – 4x2 + x + 6
3) p(x) = x4 – x3 – 7x2 + x + 6
4) p(x) = x4 + 3x3 + x2 – 3x - 2
5) p(x) = 2x3 – 11x2 + 4x + 5

Answer 1

Answer:

x3 – 3x2 – 10x + 24 Let p(x) = x3 – 3x2 – 10x + 24 Sum of all the co-efficients = 1 – 3 – 10 + 24 = 25 – 13 = 12 ≠ 0 Hence (x – 1) is not a factor. Sum of co-efficient of even powers with constant = -3 + 24 = 21 Sum of co-efficients of odd powers = 1 – 10 = – 9 21 ≠ -9 Hence (x + 1) is not a factor. p (2) = 23 – 3(22) – 10 x 2 + 24 = 8 – 12 – 20 + 24 = 32 – 32 = 0 ∴ (x – 2) is a factor. Now we use synthetic division to find other factor Thus (x – 2) (x + 3) (x – 4) are the factors. ∴ x3 – 3x2 – 10x + 24 = (x – 2) (x + 3) (x – 4)Read more on Sarthaks.com - https://www.sarthaks.com/965034/factorise-the-following-polynomial-using-synthetic-division-x-3-3x-2-10x-24