Question

If the polynomials 2x^3+ ax^2 + 3x – 5 and x^3 + x^2- 4x + a leave the same remainder,
when divided by x-2, find the value of a.​

Answer 1

Given:-

$2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $(x - 2).$

To find:-

• Value of $a.$

Answer:-

Let

• $p(x) = 2x^3 + ax^2 + 3x - 5$
• $f(x) = x^3 + x^2 - 4x + a$

Let the common remainder obtained in dividing $p(x)$ and $f(x)$ by $(x-2)$ be $r.$

Reminder theorem:

On dividing a polynomial $p(x)$ by $(x-k),$ the remainder is $p(k).$

As we are dividing by $(x-2),$ $k=2.$

For p(x):

$r = p(k)$

$\longrightarrow \: r = p(2)$

${\longrightarrow \: r = 2( {2}^{3} ) + a( {2}^{2} ) + 3(2) - 5}$

$\longrightarrow \: r = 16 + 4a +6 - 5$

$\longrightarrow \: r = 17 + 4a \quad \quad \dots(1)$

For f(x):

$r = f(k)$

$\longrightarrow \: r = f(2)$

$\longrightarrow r = ({2}^{3} ) + ( {2}^{2} ) - 4(2) + a$

$\longrightarrow r = 8 + 4 - 8 + a$

$\longrightarrow r = 4 + a \quad \quad \dots(2)$

From (1) and (2):

$17 + 4a = 4 + a$

$\longrightarrow \underline{ \underline{a = \dfrac{ - 13}{3} }}$

Answer 2

Answer:

Let the given polynomials be f(x) and g(x).

When f(x) and g(x) are divided by (x-2) they leave the same remainder.

I.e (x-2) is a factor of f(x) and g(x). It means 2 is the zero of f(x) and g(x)

So that,

f(2) = g(2)

2x³+ax²+3x-5 = x³+x²-4x+a

2(2)³+a(2)²+3(2)-5 = 2³+2²-4(2)+a

2(8)+a(4)+6-5 = 8+4-8+a

16+4a+1 = 4+a

17+4a = 4+a

4a-a = 4-17

3a = -13

a = -13/3.