If $ 2 x+3 $ and $ x+2 $ are the factors of the polynomial $ g(x)=2 x^{3}+a x^{2}+27 x+b $, find the value of the constants $a$ and $b$.

Given :

The given polynomial is $g(x) = 2x^3 + ax^2 + 27x + b$.

$2x + 3$ and $x + 2$ are the factors of the polynomial $g(x) = 2x^3 + ax^2 + 27x + b$.

To do :

We have to find the value of the constants $a$ and $b$.

Solution :

$2x + 3$ and $x + 2$ are the factors of the polynomial $g(x) = 2x^3 + ax^2 + 27x + b$.

At $x = -2$,

$g(-2) = 2(-2)^3 + a(-2)^2 + 27(-2) + b = 0$.

$2(-8)+4a-54+b=0$

$4a + b - 70 = 0$

$4a + b = 70$-----(i)

At $x = \frac{-3}{2}$,

$g(\frac{-3}{2}) = 2(\frac{-3}{2})^3 + a(\frac{-3}{2})^2 + 27(\frac{-3}{2}) + b = 0$.

$2(\frac{-27}{8})+\frac{9a}{4}-\frac{81}{2}+b=0$

$\frac{-27+9a-2(81)+4(b)}{4}=0$

$9a + 4b - 189 = 0$

$9a + 4b = 189$----(ii)

To solve the above two equations we multiply equation (i) by 4 so that $4b$ gets cancelled and we can find the value of $a$ first.

$4(4a+b) = 4(70)$

$16a+4b = 280$ -----(iii)

Now,

Equation (iii) $-$ equation (i) is,

$16a+4b = 280$ 

$-(9a+4b = 189)$

--------------------

$7a = 91$

$a = \frac{91}{7}$

$a = 13$

Substitute $a = 13$ in equation (i)

$4(13)+b = 70$

$52+b =70$

$b = 70-52$

$b = 18$.

The value of $a$ is 13 and $b$ is 18.

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Updated on: 10-Oct-2022

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