Question

Let R1 and R2 are the remainders when the polynomials
x3
+2x^2
– 5ax – 7 and x3

+ ax2

– 12x + 6 are divided

by x – 1 and x + 2 respectively. If 2R1

+ R2

= 2, then the value of a is

Answer 1

Answer:

$\text{The value of a is 2}$

Step-by-step explanation:

$\text{Given the polynomials }P(x)=x^3+2x^2-5ax-7\text{ and }P'(x)=x^3+ax^2-12x+6$

$\text{when the above polynomials are divided by x - 1 and x + 2 resp.}$

$\text{then }R_1\text{ and }R_2\text{ are remainders such that}$

$2R_1+R_2=2$

By remainder theorem

$P(1)=x^3+2x^2-5ax-7=1^3+2(1)^2-5a(1)-7=1+2-5a-7=-4-5a$

$P'(-2)=x^3+ax^2-12x+6=(-2)^3+a(-2)^2-12(-2)+6=-8+4a+24+6=22+4a$

As given

$2P(1)+P'(-2)=2$

$2(-4-5a)+22+4a=2$

$-8-10a+22+4a=2$

$6a=12$

$a=2$

Answer 2

Given:

$R_1$ and $R_2$ are remainders when the polynomials

$x^3+2x^2-5ax-7$ and $x^3+ax^2-12x+6$ are divided by x-1 and x+2 respectively.

$2R_1+R_2=2$

To find:

Value of a

Solution:

If x-1 divides the polynomial $x^3+2x^2-5ax-7$

Substitute x=1

$f(1)=1+2-5a-7$

$f(1)=-5a-4$

Then, using remainder theorem

$R_1=f(1)=-5a-4$

x+2=0

$\implies x=-2$

$f'(x)=x^3+ax^2-12x+6$

Substitute x=-2

$f'(-2)=(-2)^3+4a-12(-2)+6$

$f'(-2)=-8+4a+24+6$

$f'(-2)=22+4a$

Using remainder theorem

$R_2=22+4a$

Substitute the value

$2(-5a-4)+22+4a=2$

$-10a-8+22+4a=2$

$-6a=2-22+8$

$-6a=-12$

$a=-\frac{12}{-6}=2$