Question

please answer this ques ​

Answer 1

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,

$\leadsto{\large{\boxed{\rm{ b = - a + 4 }}}}{\longrightarrow{\text{Equation 1 }}}$

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 = $\dfrac{-16}{2}$

$\leadsto{\large{\boxed{\rm{a = - 8 }}}}$

Now substitute the value of a in equation 1

So we get ,

➜ b = - a + 4

➜ b = - ( - 8 ) + 4

➜ b = 8 + 4

$\leadsto{\large{\boxed{\rm{b = 12 }}}}$

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

$\rule{400}{6}$

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 )

$\leadsto{\large{\boxed{\tt{ r = - 4p - 10 }}}}{\longrightarrow{\text{Equation 1 }}}$

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 ($\dfrac{1}{4}$ ) + $\dfrac{5}{2}$ + r = 0

$\dfrac{ p }{4} + \dfrac{5}{2}$ + r = 0

Substitute the value of " r " from equation 1

$\longrightarrow{\tt{ \dfrac{p}{4} + \dfrac{5}{2} - 4p - 10 = 0 }}$

By taking LCM we get ,

$\longrightarrow{\tt{ \dfrac{ p + 10 - 16p - 40 }{4} = 0 }}$

$\longrightarrow{\tt{ \dfrac{ - 15p - 30 }{4} = 0 }}$

Transpose denominator 4 to the right side

$\longrightarrow{\tt{ - 15p - 30 = 0 }}$

$\longrightarrow{\tt{ - 15p = 30 }}$

$\longrightarrow{\tt{ p = \dfrac{30}{- 15}}}$

$\leadsto{\large{\boxed{\tt{ p = - 2 }}}}$

Now substitute the value of p in equation 1

➜ r = - 4p - 10 = 0

➜ r = - 4 ( - 2 ) - 10

➜ r = 8 - 10

$\leadsto{\large{\boxed{\tt{r = - 2 }}}}$

Hence,

p = -2 , r = -2

Therefore,

➜ p = r