Question

the polynomials p(x)=a^3+4x^2+3x-4 and q(x)=x^3-4x+a leave the same reminders when duvided by x-3.find the reminder when p(x) iz divided by x-2​

Answer 1

Correct Question :-

• The polynomials p(x)=ax^3 + 4x^2 + 3x - 4 and q(x) = x^3- 4x + a leave the same reminders when divided by x - 3. Find the remainder when p(x) is divided by x - 2.

Answer

Concept Used :-

Concept Used :- Remainder Theorem

• Remainder theorem in algebra states that if f(x) is a polynomial in x then the remainder on dividing f(x) by x − a is f(a).

$\large\underline{\bold{Solution :- }}$

$\rm :\longmapsto\:p(x) = {ax}^{3} + {4x}^{2} + 3x - 4$

• When p(x) is divided by (x - 3), the remainder is given by

$\rm :\longmapsto\:remainder \: =p(3)$

$\rm :\longmapsto\:remainder \: =a {(3)}^{3} + 4 \times {(3)}^{2} + 3 \times 3 - 4$

$\rm :\longmapsto\:remainder \: =27a + 36 + 9 - 4$

$\rm :\longmapsto\:remainder \: =27a + 41 - - (1)$

Now,

$\rm :\longmapsto\:q(x) = {x}^{3} - 4x + a$

• When q(x) is divided by x - 3, the remainder is given by

$\rm :\longmapsto\:remainder \: =q(3)$

$\rm :\longmapsto\:remainder \: = {(3)}^{3} - 4 \times 3 + a$

$\rm :\longmapsto\:remainder \: =27 - 12 + a$

$\rm :\longmapsto\:remainder \: =15 + a$

$\large \underline{\tt \:{ According \: to \: statement }}$

• p(x) and q(x) leaves the same remainder when divided by x - 3,

So,

$\rm :\implies\:p(3) = q(3)$

$\rm :\implies\:27a + 41 = 15 + a$

$\rm :\implies\:26a = - 26$

$\rm :\implies\:a \: = \: - 1$

So,

• p(x) can be rewritten as

$\rm :\longmapsto\:p(x) = { - x}^{3} + {4x}^{2} + 3x - 4$

Now,

• we have to find the remainder when p(x) is divided by x - 2

So,

By using Remainder Theorem,

• the Remainder when p(x) is divided by x - 2 is given by

$\rm :\longmapsto\:remainder \: =p(2)$

$\rm :\longmapsto\:remainder \: = - {(2) }^{ 3} + 4 \times {(2)}^{2} + 3 \times 2 - 4$

$\rm :\longmapsto\:remainder \: = - 8 + 16 + 6 - 4$

$\rm :\longmapsto\:remainder \: =10$