Question

find the value of k when x-1 is the factor of p(x)=kx^2-2x+1

Answer 1

p(x) = kx ^2 - 2x + 1

g(x) = x - 1

by remainder theorem

x = 1

p(1) = 0

p(1) = k(1)^2 - 2(1) + 1

0 = k - 2 + 1

0 = k - 1

1 = k

Answer 2

Value of $\bf \red{k=1}$, if (x-1) is a factor of p(x).

Given:

• A polynomial $p(x) = k {x}^{2} - 2x + 1$
• $(x - 1)$ is a factor .

Solution:

Theorem/Concept to be used:

Remainder Theorem: It states that if (x-a) is a factor of p(x), then p(a)=0.

Step 1:

Find the value of x from the factor.

$x - 1 = 0$

or

$\bf x = 1$

Step 2:

Put x=1 in p(x).

From remainder theorem it is clear that p(1)=0.

So,

$k( {1)}^{2} - 2(1) + 1 = 0$

or

$k - 2 + 1 = 0$

or

$k - 1 = 0$

or

$\bf k = 1$

Thus,

Value of k=1, if (x-1) is a factor of p(x).

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