Question

for what value of k,the HCF of
$x ^{2} + x + (5k - 1)$
and
$x {}^{2} - 6x + (3k + 11)$
is
$(x - 2)$


​

Answer 1

Question:

For what value of "k" , the HCF of the polynomials x^2 + x + (5k-1) and

x^2 - 6x + (3k+11) is (x-2).

Answer:

k = -1

Note:

If (x-a) is a factor of any polynomial p(x) then x = a its zero and p(a) = 0.

Solution:

Let ,

p(x) = x^2 + x + (5k-1)

g(x) = x^2 - 6x + (3k+11)

It is given that, the HCF of the polynomials x^2 + x + (5k-1) and

x^2 - 6x + (3k+11) is (x-2).

Thus,

It is clear that, (x - 2) is the common factor of the polynomials

p(x) = x^2 + x + (5k-1) and

g(x) = x^2 - 6x + (3k+11).

Hence,

x = 2 is the common zero of the polynomials p(x) = x^2 + x + (5k-1) and

g(x) = x^2 - 6x + (3k+11).

Thus,

We have , p(2) = 0 and g(2) = 0.

When p(2) = 0 ,then;

=> (2)^2 + 2 + (5k-1) = 0

=> 4 + 2 + 5k - 1 = 0

=> 5k + 5 = 0

=> 5k = - 5

=> k = - 5/5

=> k = -1

When g(2) = 0 ,then;

=> (2)^2 - 6•2 + (3k + 11) = 0

=> 4 - 12 + 3k + 11 = 0

=> 3k + 3 = 0

=> 3k = - 3

=> k = - 3/3

=> k = -1

Hence,

The appropriate value of "k" is (-1).

Answer 2

AnswEr :

• ( x – 2 ) is Highest Common Factor of

{x² + x + (5k – 1)} and {x² – 6x + (3k + 11)}

⋆ First from the Factor :

» ( x – 2 ) = 0

» x = 2

⋆ Now we will put the Value of x = 2 in Different Equation to find the Value of K :

Let p(x) = x² + x + (5k – 1)

⇒ p(x) = 0

⇒ x² + x + (5k – 1) = 0

⇒ ( 2 )² + 2 + (5k – 1) = 0 [ x = 2 ]

⇒ 4 + 2 + 5k – 1 = 0

⇒ 5 + 5k = 0

⇒ 5k = –5

• Dividing Both Term by 5

⇒ k = –1

Let g(x) = x² – 6x + (3k + 11)

⇒ g(x) = 0

⇒ x² – 6x + (3k + 11) = 0

⇒ ( 2 )² – 6*2 + (3k + 11) = 0 [ x = 2 ]

⇒ 4 – 12 + 3k + 11 = 0

⇒ 3 + 3k = 0

⇒ 3k = –3

• Dividing Both Term by 3

⇒ k = –1

჻ Value of k = –1 will satisfy the HCF (x – 2) for both the Equations.