please tell - how do we do ques 4 and 5 ?
Attachments:

Answers
Answered by
1
g (x)=0
x=2
putting the values, we get the value of k
x=2
putting the values, we get the value of k
Attachments:

lonelypark09:
why did we put 2k=22?
Answered by
1
(Q.4.)
I assume you know the Division algorithm, i.e. dividend=divisor*quotient+remainder.
if p(x) is completely divisible by x-2, the remainder will be 0.
So, divide (3x²-kx+10) by (x-2) in such a way that the remainder is zero. From that you will find that the value of k is 1.
(Q.5.)
You probably know that a²-b²=(a+b)(a-b)
In this question, let a = x²+y²-z² and b = x²-y²+z².
So, (x²+y²-z²)² - (x²-y²+z²)² = (x²+y²-z²+x²-y²+z²)(x²+y²-z²-x²+y²-z²)
=2x²×(2y²-2z²)
=4x²y²-4x²z²
I assume you know the Division algorithm, i.e. dividend=divisor*quotient+remainder.
if p(x) is completely divisible by x-2, the remainder will be 0.
So, divide (3x²-kx+10) by (x-2) in such a way that the remainder is zero. From that you will find that the value of k is 1.
(Q.5.)
You probably know that a²-b²=(a+b)(a-b)
In this question, let a = x²+y²-z² and b = x²-y²+z².
So, (x²+y²-z²)² - (x²-y²+z²)² = (x²+y²-z²+x²-y²+z²)(x²+y²-z²-x²+y²-z²)
=2x²×(2y²-2z²)
=4x²y²-4x²z²
Similar questions