Question

The polynomials $a{x}^3+3x^2$ - 3 and $2x^3$ - 5x + a when divided by (x-4) leave the remainders $R_1$ and $R_2$ respectively. Find the values of a in each case of the following cases, if

(i) $R_1=R_2$

(ii) $R_1+R_2=0$

(iii) $2R_1-R_2=0$

Answer 1

Let f(x) = $a{x}^3+3x^2$ - 3

g(x) = $2x^3$ - 5x + a

and divisor q(x) = x - 4

x - 4 = 0 $\Rightarrow$ x = 4

Now remainder in each case,

$R_1$ = f(4) = $a(4{)}^3+3(4{)}^2$ - 3 = 64a + 48-3

= 64a + 45

and $R_2$ = g(4) = $2(4{)}^3$ - 5(4) + a = 2 $\times$ 64

- 20 + a

= 128 - 20 + a = 108 + a

Now,

(i) If $R_1=R_2$, then

64a + 45 = 108 + a $\Rightarrow$ 64a - a = 108 - 45

$\Rightarrow$ 63a = 63 $\Rightarrow a=\frac{63}{63}=1$

a = 1

(ii) If $R_1+R_2$ = 0

$\Rightarrow$ 64a + 45 + 108 + a = 0 $\Rightarrow$ 65a + 153 = 0

$\Rightarrow$ 65a = - 153 $\Rightarrow a=\frac{-153}{65}$

$\therefore a=\frac{-153}{65}$

(iii) $2R_1-R_2=0$ $\Rightarrow$ 2(64a+ 45) - 108 + a = 0

$\Rightarrow$ 128a + 90 - 108 - a = 0

$\Rightarrow$ 127a - 18 = 0$\Rightarrow$ 127 a = 18

$\Rightarrow a=\frac{18}{127}$

$\therefore a=\frac{18}{127}$