Question

when a polynomial 2x³ + 3x²+ax +b is divided by ( x-2 ) leaves remainder 2, and ( x + 2 ) leaves remainder -2 find a and b ​

Answer 1

Answer:

Step-by-step explanation:

Given polynomial is  $\sf{2{x}^{3}+3{x}^{2}+ax+b}[\tex]</p><p>→ When it is divided by  [tex]\left(x-2\right)$, leaves remainder 2

Then,  $\sf{2{x}^{3}+3{x}^{2}+ax+b-2}[\tex]  is divisible by  [tex]\left(x-2\right)$

So,

$\left(x-2\right)$  is a factor of  $\sf{2{x}^{3}+3{x}^{2}+ax+b-2}[\tex].</p><p>So,</p><p>[tex]\sf{2\left(2\right)^3+3\left(2\right)^2+a\left(2\right)+b-2=0}$

$\sf{\implies16+12+2a+b-2=0}$

$\sf{\implies2a+b+26=0\,\,\,\,\,\,\,...(1)}$

→ When it is divided by  $\left(x+2\right)$, leaves remainder -2

Then,  $\sf{2{x}^{3}+3{x}^{2}+ax+b+2}[\tex]  is divisible by  [tex]\left(x+2\right)$

So,

$\left(x+2\right)$  is a factor of  $\sf{2{x}^{3}+3{x}^{2}+ax+b+2}[\tex].</p><p>So,</p><p>[tex]\sf{2\left(-2\right)^3+3\left(-2\right)^2+a\left(-2\right)+b+2=0}$

$\sf{\implies\,-16+12-2a+b+2=0}$

$\sf{\implies\,-2a+b-2=0\,\,\,\,\,\,\,\,...(2)}$

Adding (1) and (2), we get,

$\sf{\implies\,2b+24=0}$

$\sf{\implies\,\boxed{b=-12}}$

Putting the value of b in (1),

$\sf{\implies2a-12+26=0}$

$\sf{\implies2a+14=0}$

$\sf{\implies\,\boxed{a=-7}}$