Question

Find the value of 'a' and 'b' so that the polynomial x3-ax2-13xb has (x-1) and (x3) as the factors.

Answer 1

HEY!!

p(x) = x3 - ax2 - 13x + b

x-1 = 0
x = 1
p(1) = 13 - a(1)2 - 13(1) + b = 0
p(1) = 1-a-13+b = 0
p(1) = -a-12+b = 0

x+3 = 0
x = -3
p(-3) = (-3)3 - a(-3)2 - 13(-3) + b = 0
p(-3) = -27 - 9a + 39 + b = 0
p(-3) = - 9a + 12 + b = 0

a + 12 - b = 0 (1)
-9a + 12 + b = 0 (2)

Add (1) and (2) to get:

-8a + 24 = 0
8(-a + 3) = 0
-a + 3 = 0
a = 3

-3-12+b = 0
-15+b = 0
b = 15