Question

If (x—1) is a factor of x³—kx²+11x—6, then the value of k should be

Answer 1

(x-1) is a factor
x-1=0
x=1
p(x) = x³—kx²+11x—6
p(1) = (1)³—k(1)²+11(1)—6
= 1+k(1)+11(1)—6
= 1+11-6+k
= 12-6+k
= 6+k
-k= 6
k = -6

Answer 2

Answer:

The value of k = 6

Step-by-step explanation:

Given,

(x—1) is a factor of x³—kx²+11x—6

To find,

The value of 'k'

Recall the concept

Factor theorem:

p(x) is a polynomial and the linear polynomial (x-a) is a factor of f(x) then p(a) = 0

Solution:

Let p(x) =  x³- kx²+11x - 6

and x -a = x-1

Then by factor theorem we have

p(1) = 0

1³ - k×1²+11×1 - 6 = 0

1 - k +11 - 6 = 0

-k+6 = 0

k = 6

∴ The value of k = 6

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