please answer this ques

Answers
Question - 1
If 2x³ + ax² + bx - 6 has x - 1 as a factor and leaves a remainder 2 when divided by x - 2 , find the value of " a " and " b " ?
Answer :-
P ( x ) = 2x³ + ax² + bx - 6
Given that :-
x - 1 is a factor
So,
Let, x - 1 = 0
=> x = 1
Now substitute this value in place of x in p(x)
Hence,
➜ P ( 1 ) = 2 ( 1 )³ + a ( 1 )² + b ( 1 ) - 6
➜ 2 ( 1 )³ + a ( 1 )² + b ( 1 ) - 6 = 0
➜ 2 ( 1 ) + a ( 1 ) + b - 6 = 0
➜ 2 + a + b - 6 = 0
➜ a + b - 4 = 0
(Transpose a , - 4 to the right side )
So,
Similarly,
It is also given that :-
When p ( x ) is divided by x - 2 it leaves remainder 2
So,
Let , x - 2 = 0
=> x = 2
Now, substitute this value in place of x in p ( x )
So,
➜ P ( 2 ) = 2 ( 2 )³ + a ( 2 )² + b ( 2 ) - 6 = 2
Transposing 2 from right side to left side
➜ 2 ( 2 )³ + a ( 2 )² + b ( 2 ) - 6 - 2 = 0
➜ 2 ( 8 ) + a ( 4 ) + 2b - 8 = 0
➜ 16 + 4a + 2b - 8 = 0
➜ 8 + 4a + 2b = 0
Substitute the value of " b " from equation 1
➜ 8 + 4a + 2 ( - a + 4 ) = 0
➜ 8 + 4a - 2a + 8 = 0
➜ 16 + 2a = 0
➜ 2a = - 16
➜ a =
Now substitute the value of a in equation 1
So we get ,
➜ b = - a + 4
➜ b = - ( - 8 ) + 4
➜ b = 8 + 4
Verification :-
p ( x ) = 2x³ + ax² + bx - 6
Substitute the values of a and b in p(x)
P ( x ) = 2x³ + ( - 8 )x² + (12)x - 6
P( x ) = 2x³ - 8x² + 12x - 6
x - 1 is a factor
So,
x - 1 = 0
x = 1
When this value is substituted in place of x in p (x ) then the remainder should be zero if it is zero then the values of a and b are correct
So,
➜ p ( 1 ) = 2 ( 1 )³ - 8 ( 1 )² + 12 ( 1 ) - 6
➜ 2 ( 1 ) - 8 ( 1 ) + 12 - 6 = 0
➜ 2 - 8 + 12 - 6 = 0
➜ 14 - 14 = 0
➜ 0 = 0
L.H.S. = R.H.S.
Hence verified
Question - 2
If both ( x - 2 ) and ( x - ½ ) are factors of px² + 5x + r , show that p = r
Answer :-
P( x ) = px² + 5x + r
Given that :-
( x - 2 ) is a factor
So,
Let , x - 2 = 0
=> x = 2
➜ p ( 2 ) = p ( 2 )² + 5 ( 2 ) + r
➜ P ( 4 ) + 5 ( 2 ) + r = 0
➜ 4p + 10 + r = 0
(Transposing 4p and 10 to the right side )
Similarly,
It is also given that :-
( x - ½ ) is also a factor of p ( x )
So,
Let , x - ½ = 0
x = ½
➜ p ( ½ ) =
➜ p ( ½ )² + 5 ( ½ ) + r = 0
➜ p ( ) +
+ r = 0
+ r = 0
Substitute the value of " r " from equation 1
By taking LCM we get ,
Transpose denominator 4 to the right side
Now substitute the value of p in equation 1
➜ r = - 4p - 10 = 0
➜ r = - 4 ( - 2 ) - 10
➜ r = 8 - 10
Hence,
p = -2 , r = -2