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
iii. 2R1 + R2= 0

Answer 1

Answer:

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=0The value of a is

9

16

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 }Given that the polynomials P(x)=x

3

+2x

2

−5ax−7 and x

3

+ax

2

−12x+6 when divided by x+1 and x−2 respectively leaves the remainder R1 and R2

we have to find the value of a

By remainder theorem,

P(-1)=R_1P(−1)=R

1

for first polynomial

(-1)^3+2(-1)^2-5a(-1)-7=R_1(−1)

3

+2(−1)

2

−5a(−1)−7=R

1

5a-6=R_15a−6=R

1

P(2)=R_2P(2)=R

2

for second polynomial.

2^3+a(2)^2-12(2)+6=R_22

3

+a(2)

2

−12(2)+6=R

2

4a-10=R_24a−10=R

2

Case 1: R_1=R_2R

1

=R

2

5a-6=4a-105a−6=4a−10

a=-4a=−4

\text{The value of a is -4 when }R_1=R_2The value of a is -4 when R

1

=R

2

Case 2: R_1+R_2=0R

1

+R

2

=0

5a-6+4a-10=05a−6+4a−10=0

9a=169a=16

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

9

16

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

9

16

when R

1

+R

2

=0

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