Question

We get the same remainder if polynomial​

Answer 1

Question :

We get the same remainders if the polynomials x^3 + x^2 - 4x + a and

2x^3 + ax^2 + 3x - 3 are divided by

(x-2). Then find the value of "a".

Answer :

a = - 5

Note :

Remainder theorem:

If polynomial p(x) is divided by (x-a) , then the remainder is given by ;

r = p(a).

Solution :

Case(1),

When the polynomial

x^3 + x^2 - 4x + a

is divided by (x-2) ,then the reminder will be given by;

=> r1 = x^3 + x^2 - 4x + a

=> r1 = (2)^3 + (2)^2 - 4(2) + a

=> r1 = 8 + 4 - 8 + a

=> r1 = a + 4

Case(2),

When the polynomial

2x^3 + ax^2 + 3x - 3

is divided by (x-2) ,then the reminder will be given by;

=> r2 = 2x^3 + ax^2 + 3x - 3

=> r2 = 2(2)^3 + a(2)^2 + 3(2) - 3

=> r2 = 16 + 4a + 6 - 3

=> r2 = 4a + 19

According to the question;

We get the same remainders if the polynomials (x^3 + x^2 - 4x + a) and

(2x^3 + ax^2 + 3x - 3) are divided by

(x-2).

Thus;

=> r1 = r2

=> a + 4 = 4a + 19

=> 4a - a = 4 - 19

=> 3a = - 15

=> a = -15/3

=> a = - 5

Hence;

The required value of "a" is (- 5 ).

Answer 2

Answer:-

$\implies \: \boxed{ a = - 5}$

Step - by - step explanation:-

Condition used:-

Here ,we used Reminder theorem .

Reminder theorem :-

If a polynomial P(x) is divided by linear polynomial (x - a) then,the reminder is P(a) .

Solution:-

Let,

$</em><em>Q</em><em>(x) = {x}^{3} + {x}^{2} - 4x + a \\ \\ and \\ \\ </em><em>P</em><em>(x) = 2 {x}^{3} + a {x}^{2} + 3x - 3$

According to the question,

→ P (x) and Q(x) is divisible by ( x - 2 ) .

Hence,

According to the given condition ↓

Q(2) and P(2) will reminder →

Therefore,

$\: Q(2) = {(2)}^{3} + {(2)}^{2} - 4 \times 2 + a \\ \\ Q(2) = \: 8 + 4 - 8 + a \\ \\ Q(2) = 4 + a$

Now for P(x) ↓

$P(2) = 2 \times {(2)}^{3} + a \times {(2)}^{2} + 3 \times 2 - 3 \\ \\ P(2) = 16 + 4a + 6 - 3 \\ \\ P(2) = 4a + 19$

According to the question,

P(x) and Q(x) both are divisible by (x -2)

Hence ,

Their reminders be equal .

$\implies \: Q(2) = P(2) \\ \\ \implies \: 4 + a = 4a + 19 \\ \\ \implies \: 3a = - 15 \\ \\ \implies \: \boxed{ a = - 5}$ .