Question

The polynomials x3 + 2x2 -5ax -7 and x3 + ax2 –12x +6 when divided by x + 1 and x – 2 respectively, leave remainders R1 and R2 respectively. Find the value of a in each of the following cases: i)R1 = R2 ii). R1 + R2 =0

Answer 1

.............by using remainder theorem in first case a= -10. and in second case a= 22/9

Answer 2

Answer:

$\text{The value of a is -4 when }R_1=R_2$

$\text{The value of a is }\frac{16}{9}\text{ when }R_1+ R_2=0$

Step-by-step explanation:

$\text{Given that the polynomials }P(x)=x^3+2x^2-5ax-7\text{ and }x^3+ax^2-12x+6\text{ when divided by }x+1\text{ and }x-2\text{ respectively leaves the remainder R1 and R2 }$

we have to find the value of a

By remainder theorem,

$P(-1)=R_1$ for first polynomial

$(-1)^3+2(-1)^2-5a(-1)-7=R_1$

$5a-6=R_1$

$P(2)=R_2$ for second polynomial.

$2^3+a(2)^2-12(2)+6=R_2$

$4a-10=R_2$

Case 1: $R_1=R_2$

$5a-6=4a-10$

$a=-4$

$\text{The value of a is -4 when }R_1=R_2$

Case 2: $R_1+R_2=0$

$5a-6+4a-10=0$

$9a=16$

$a=\frac{16}{9}$

$\text{The value of a is }\frac{16}{9}\text{ when }R_1+R_2=0$