Ratio and proportion

Trignometry

Remainder Theorem

Show that (x - 1) is a factor of x³ – 7x ² + 14x – 8. Hence completely factorize

                                                           (ICSE 2007)

Answer:

Let: F(x) = x³ – 7x ² + 14x – 8

Since (x - 1) is a factor of F(x), therefore, according to Remainder theorem,

we have, F (1) = 0

Now, F (1) = (1)³ - 7(1)² + 14(1) – 8

Or, F(1) =1 – 7 +14 - 8 = 0

So, (x – 1) is a factor.

Now we will divide (x³ – 7x ² + 14x – 8)/(x – 1)

(x - 1) // x³ – 7x ² + 14x – 8 //  x ² - 6x + 8

             x³ - x ² 

               - 6 x ² + 14x

              - 6 x ² +  6x

                            8x – 8

                           8x – 8

                           0     0  

(x ² - 6x +8) is the other factor of F(x). We will further simplify, thus:

x ² - 6x +8 = x ² - 4x - 2x + 8 = x(x – 4) – 2(x – 4) = (x – 2)(x – 4)

So, x³ – 7x ² + 14x – 8  = (x – 1 )(x – 2)(x – 4)

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