Question

If P(x) = 2x³ + 2x² + bx - 6 leaves remainder 36, when divided by (x-3) , find the value of b and hence factorise P(x).

Answer 1

${\huge {\mathfrak {Answer :-}}}$

We have, p(x) = 2x^3 + 2x^2 + bx - 6,

And when, p(x) is divided by (x - 3) it leaves a remainder 36.

So, by using

${\mathfrak {\pink {-:Remainder \: theorem :-}}}$

p(3) = 2(3)^3 + 2(3)^2 + b(3) - 6 = 36

Or, 54 + 18 + 3b - 6 = 36

Or, 3b = 36 - 66 = - 30

Or, b = - 30/3 = - 10

➡️ $<b>Hence, the value of p is (- 10).</b>$

Now, we need to factorise p(x)

So, we have

p(x) = 2x^3 + 2x^2 + (- 10)x - 6 - 36

= 2x^3 + 2x^2 - 10x - 42

= 2x^3 - 6x^2 + 8x^2 - 24x + 14x - 42

= 2x^2(x - 3) + 8x(x - 3) + 14(x - 3)

= (x - 3)(2x^2 + 8x + 14)

= 2(x - 3)(x^2 + 4x + 7)

$<b>$ That's it..

Answer 2

Answer :-

$\boxed{\textsf{b=-10}}$

$\boxed{\boxed{\bf{2 [( x^2+4x+7 )(x-3 )+18]}}}$

Given polynomial P(x) = 2 x³ + 2 x² + b x - 6 .

When divided by ( x-3 ) P(x) leaves a remainder 36 .

According to the remainder theorem :

If ( x - a ) leaves a remainder y then f (a) = y

Applying the above theorem :

P(3) = 36

P(x) = 2 x³ + 2 x² + b x - 6

⇒ P(3) = 2(3)³ + 2(2)² + b(2) - 6

⇒ 36 = 2 × 27 + 2 × 4 + 2b - 6

⇒ 54 + 8 + 2b - 6 = 36

⇒ 56 + 2b = 36

⇒ 2b = 36 - 56

⇒ 2b = - 20

⇒ b = -20/2

⇒ b = - 10

The value of b is - 10 .

So the original equation becomes 2 x³ + 2 x² - 10 x - 6

Since the function P(x) leaves remainder 36 when divided by x - 3 :-

P(x) - 36 is a factor of x - 3 .

⇒ 2 x³ + 2 x² - 10 x - 6 - 36

⇒ 2 x³ + 2 x² - 10 x - 42 is a factor of x - 3

⇒ 2 ( x³ + x² - 5 x - 21 )

Split these terms :-

⇒ 2 ( x³ - 3 x² + 4 x² - 12 x + 7 x - 21 )

⇒ 2 ( x²( x - 3 ) + 4 x ( x - 3 ) + 7 ( x - 3 )

⇒ 2 ( x² + 4 x + 7 )( x - 3 )

This was done by subtracting 36 from it .

Hence P(x) - 36 = 2 ( x² + 4 x + 7 )( x - 3 )

⇒ P(x) = 2 ( x² + 4 x + 7 )( x - 3 ) + 36

⇒ P(x) = 2 [ ( x² + 4 x + 7 )( x - 3 ) + 18 ]

Hence P(x) is factorized .