Question

If the polynomials x3 + ax2 +5 and x3-2x2+ a are divided by (x + 2) leave the same
remainder, find the value of a.
​

Answer 1

Answer :

a = -13/3

Note :

★ Remainder theorem : If a polynomial p(x) is divided by (x - c) , then the remainder obtained is given as R = p(c) .

★ Factor theorem :

If the remainder obtained on dividing a polynomial p(x) by (x - c) is zero , ie. if R = p(c) = 0 , then (x - c) is a factor of the polynomial p(x) .

If (x - c) is a factor of the polynomial p(x) , then the remainder obtained on dividing the polynomial p(x) by (x - c) is zero , ie. R = p(c) = 0 .

Solution :

Let

f(x) = x³ + ax² + 5

g(x) = x³ - 2x² + a

Also ,

If x + 2 = 0 , then x = -2 .

Thus ,

Accounting to the remainder theorem , if f(x) is divided by (x + 2) , then the remainder R will be f(-2) .

Thus ,

=> R = f(-2)

=> R = (-2)³ + a(-2)² + 5

=> R = -8 + 4a + 5

=> R = 4a - 3

Also ,

If g(x) is divided by (x + 2) , then the remainder R' will be g(-2) .

Thus ,

=> R' = g(-2)

=> R' = (-2)³ - 2(-2)² + a

=> R' = -8 - 8 + a

=> R' = a - 16

But

According to the question , if f(x) and g(x) is divided by (x + 2) leave the same remainder .

Thus ,

=> R = R'

=> 4a - 3 = a - 16

=> 4a - a = 3 - 16

=> 3a = -13

=> a = -13/3

Hence a = -13/3

Answer 2

Answer:

Given:-

• Polynomials x³+ ax²+5 and x²-2x²+ a are divided by (x + 2) leave the same remainder

Find:-

• Value of A

Solution:-

The polynomials are divided with x+2

So, x = -2

${ \sf{p(x)= {x}^{3} + {ax}^{2} +5 }}$

${ \to{ \sf{p( - 2) = {( - 2)}^{3} + a {( - 2)}^{2} + 5 }}}$

${ \to{ \sf{p( - 2) = - 8 + a \times 4 + 5}}}$

${ \to{ \sf{p( - 2) = 4a - 8 + 5}}}$

${ \to{ \sf{p( - 2) = 4a - 3}}}$

${ \sf{q( x) = {x}^{3} - 2 {x}^{2} + a }}$

${ \to{ \sf{q( - 2) = {( - 2)}^{3} - 2 {( - 2)}^{2} + a }}}$

${ \to \sf{q( - 2) = - 8 - 8 + a}}$

${ \to{ \sf{q( - 2) = a - 16}}}$

According to the question

P(x) = q(x)

4a - 3 = a - 16

$4a - a = - 16 + 3$

$3a = - 13$

$a = \frac{ - 13}{3}$

Therefore the value of a is -13/3