Question

If the polynomials x^3 + ax^2 + 3x + 5x 3 +ax 2 +3x+5 and x^3 + 2x^2 + x + 2ax 3 +2x 2 +x+2a leave the same remainder when divided by (x-1)(x−1), then the value of aa is:​

Answer 1

Answer:

a=5

Step-by-step explanation:

divide both

put a=5 and get remainder 14vin both p(x)and q(x)

Answer 2

Remainder Theorem

Given:

Polynomial $x^3 + ax^2 + 3x + 5$ and $x^3+2x^2+x+2a$ are given.

These are divide by  $(x-1)$.

To find:

Value of a .

Explanation:

When a polynomial a(x) is divided by a linear polynomial b(x) whose zero is x = k, the remainder is given by r = a(k).Let the given polynomials be f(x) and g(x).

When f(x) and g(x) are divided by $(x -1)$ they leave the same remainder.

i.e. $(x -1)$ is a factor of f(x) and g(x). It means 1 is the zero of f(x) and g(x)

So that,

$f(2) = g(2)$

$x^3 + ax^2 + 3x + 5 = x^3 + 2x^{2} +x + 2a\\\\1^3+a(1^2)+3(1)+5=1^3+2(1^2)+1+2a\\\\1+a+3+5=1+2+1+2a\\\\5=a$

Hence the value of a is 5.