Question

Let ry and r2 be the remainders when the polynomials P(x)= x,^3+ x² – 5kx q(x)= x^3+ kx^2– 12x +6 are divided by x +1 and x – 2 respectively. Find the value of k if r1-r2=0​

Answer 1

Answer:

Correct option is

A

2

Let p(x)=x

3

+2x

2

−5ax−7

and q(x)=x

3

+ax

2

−12x+6 be the given polynomials,

Now, R

1

= Remainder when p(x) is divided by x+1.

⇒R

1

=p(−1)

⇒R

1

=(−1)

3

+2(−1)

2

−5a(−1)−7[∵p(x)=x

2

+2x

2

−5ax−7]

⇒R

1

=−1+2+5a−7

⇒R

1

=5a−6

And R

2

= Remainder when q(x) is divided by x-2

⇒R

1

=q(2)

⇒R

2

=(2)

3

+a×2

2

−12×2+6[∵q(x)=x

2

+ax

2

−12x−6]

⇒R

2

=8+4a−24+6

⇒R

2

=4a−10

Substituting the values of R

1

and R

2

in 2R

1

+R

2

=6, we get

⇒2(5a−6)+(4a−10)=6

⇒10a−12+4a−10=6

⇒14a−22=6

⇒14a−28=0

⇒a=2

Step-by-step explanation:

hope it's helpful