Remainder Theorem

1. Given that (x - 2) is a factor of 2x³ – x ² – px – 2. Determine the values of “p.         

                                                                                      (ICSE 2008)

Answer:

Let: F(x) = 2x³ – x ² – px – 2

Since (x - 2) is a factor of F(x), therefore, according to Remainder theorem, we have, F (2) = 0

Now, F (2) = 2(2)³ - (2)² - p(2) – 2 = 0

Or, 16 - 4 -2p – 2 = 0

Or, 2p = 10 

Or, p = 5  Answer

2.     Given that (x - 2) is a factor of 2x³ – x ² – px – 2. With the              value of p factorize the expression completely.         

                                                                                   (ICSE 2008)

Answer:

Since (x - 2) is a factor of 2x³ – x ² – 5x – 2.

The other factor has to be a trinomial of the form ax² + bx + c

Therefore (x-2)( ax² + bx + c) = 2x³ – x ² – 5x – 2.

Or, ax³ + (b -2a)x² – (c -2b)x – 2c = 2x³ – x ² – 5x – 2

Comparing the coefficients of individual power of x,  

We have a = 2,

(b – 2a) = -1

Or, b = 3

And -2c = -2

Or, c = 1

So, ax² + bx + c = 2x² + 3x +1 = 2x² + 2x + x +1= 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1)

Therefore, 2x³ – x ² – 5x – 2 = (x-2)(2x + 1)(x + 1)

Remainder Theorem

Use the Remainder Theorem to factorize the following expression:

 2x³ +x² - 13x + 6.               (ICSE 2010)

Answer:

Substitute 1,2 and 3 to check if F(x) =  2x³ +x² - 13x + 6 = 0

We find that for x = 2, F(2) = 0

So, (x -2) is a factor.

The other factor is of the form ax² + bx + c, because F(x) is a trinomial.

Therefore (x-2)( ax² + bx + c) = 2x³ +x² - 13x + 6

Or, ax³ + (b -2a)x² – (c -2b)x – 2c =2x³ +x² - 13x + 6

Comparing the coefficients of the individual powers of x on LHS and RHS,

We have a = 2,

(b – 2a) = 1

Or, b = 5

And -2c = 6

Or, c = -3

So, ax² + bx + c = 2x² + 5x – 3 = 2x² + 6x – x – 3= 2x(x + 3) – 1(x + 3) = (2x – 1)(x + 3)

Therefore, 2x³ +x² - 13x + 6 = (x-2)(2x – 1)(x + 3)  answer

Remainder Theorem

When divided by (x – 3) the polynomial x³ - px² + x + 6 and 2x³- x² – (p+3) x – 6   leaves the same remainder, then find the value of p?                   ICSE 2010

Answer:

Let: F(x) = x³ - px² + x + 6

And G(x) = 2x³- x² – (p+3) x – 6   

Let the Remainder be y

Therefore, (F(x) – y) and (G(x) – y) are completely divisible by (x – 3)

Therefore, F(3) – y = 0 and G(3) – y = 0, as per the remainder theorem.

On substitution in ( F(3) – y), we get ;

27 - 9p + 3 + 6 - y = 0 

Or, 9p + y = 36    ----------------------------------(1)

On substitution in (G(3) – y) we get  ;

54 – 9 – (p + 3)3 – 6 – y = 0   

Or, 3p + y = 30    ----------------------------------- (2)

Equating (1) and (2), we get

6p = 6

Or, p = 1

Remainder theorem

Given that (x + 2) and (x + 3) are factors of 2x³ +ax² + 7x – b. Determine the values of “a” and “b.           (ICSE 2009)

Answer:

Let: F(x) = 2x³ +ax² + 7x – b

Since (x + 2) and (x + 3) are factors of F(x), therefore, according to Remainder theorem, we have, F (-2) = F (-3) = 0

Now, F (-2) = 2(-2)³ +a (-2)² + 7(-2) – b = 0

Or,-16 + 4a -14 – b = 0

Or, 4a –b = 30 -----------------(1)

Similarly, F (-3) = 2(-3)³ +a (-3)² + 7(-3) – b = 0

Or, F (-3) = -54 +9a – 21 – b = 0

Or, 9a – b = 75 ------------------ (2)

Solving equation (1) and (2), we get

-5a = -45

Or a = 9

Therefore, b = 9x9 – 33 =48.

Answers: a = 9, b = 48

Remainder theorem

If the polynomial ax³ +4x² + 3x – 4 and x³ - 4x + a  leaves the same remainder when divided by (x – 3), then find the value of a?

Answer:

Let: F(x) = ax³ +4x² + 3x – 4

And G(x) = x³ - 4x + a  

Let the Remainder be y

Now, (F(x) – y) and (G(x) – y) are completely divisible by (x – 3)

Therefore, F(3) – y = 0 and G(3) – y = 0 and as per the remainder theorem.

On substitution we get F(3) – y ;

27a + 41 –y  = 0  ----------------------------------(1)

On substitution we get G(3) – y ;

15 + a – y = 0   -----------------------------------(2)

Equating (1) and (2), we get

27a + 41 = 15 + a

Or, 26a = 26

Or, a = 1

Remainder theorem

Without actual division, 

prove that 2x⁴ -5x³ +2x² –x +2 is divisible by x² – 3x + 2

Answer:

Let, F(x) = 2x⁴ -5x³ +2x² –x +2

And G(x) = x² – 3x + 2

Or, G(x) = (x – 2)(x – 1)

If F(x) is completely divisible by G(x), then( x - 2) and (x - 1) must be a factor of F(x).

Now, as per remainder theorem, F(2) and F(1) should be = 0, if they are completely divisible

So, checking if F(2) and F(1) = 0,

We find F(1) = 2 – 5 + 2 – 1 + 2 = 0

Also F(2) = 32 – 40 + 8 – 2 + 2 = 0

So, F(x) is completely divisible by G(x)

The polynomial x⁴ - 2x³ +3x² –ax + 3a - 7 when divided by x+ 1 leaves the remainder 19. Find the values of  a. 

Answer:

Let: F(x) = x⁴ -2x³ +3x² –ax + 3a – 7

Given,  F(x) – 19 is completely divisible by (x + 1)

Therefore, F(-1) – 19 = 0, as per the remainder theorem.

F(-1) = 4 + 2 + 3 + a + 3a – 7 = 2 + 4a

Therefore, F(-1) – 19 = 2 + 4a – 19 = 0

Or, a = 17/4

 

Problems on Remainder theorem

If  2x³ +ax² - 13x + b is divisible by x² -x - 6, then find the value of “a” and “b”.

Answer:

Factorizing x² - x -6 we get,

= x² - 3x + 2x -6

= x(x - 3) + 2(x - 3)

= (x- 3)(x + 2)

if F(x) = 2x³ +ax² - 13x + b

Now according to remainder theorem,  F(3) = 0 and F(-2) = 0 for complete divisibility.

Substituting x = 3 we get

9a + b + 15 = 0     -------------(1)

Similarly substituting x = -2, we get

4a + b + 10 = 0  ---------(2)

Solving Equation 1 and 2 we get,

“a” = -1

“b” = -6