Question

The polynomials $a{x}^3+3x^2-3$ and $2x^3-5x+a$ when divided by (x - 4) leaves remainder $R_1$ & $R_2$ such that $2R_1-R_2=0$

• A. $-\frac{18}{127}$
• B. $\frac{18}{127}$
• C. $\frac{17}{127}$
• D. $-\frac{17}{127}$

Answer 1

The correct option is B $\frac{18}{127}$

According to remainder theorem .If p(x)is divided by (x-4),then R1 will be p(4)
=$a{x}^3+3x^2-3=(4{)}^{3}a+3(4{)}^2-3=64a+48-3=64a+45--------R_1$
According to remainder theorem .If p(x)is divided by (x-4),then R2 will be p(4)
$2x^3-5x+a=2(4{)}^3-5(4)+a=128-20+a=108+a--------------R_2$
Then, $2\left(R1\right)-R2=0$
$Or,2(64a+45)-108-a=0$
$Or,128a+90-108-a=0$
$Or,127a=18$
$Or,a=\frac{18}{127}$