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
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 valie of a -4 when R

1

=R

2

text the value of a is fraction {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-7text and x^3+ax^2-12x+6 text when divided by x+1 text and x-2text respectively leaves the remainder R1 and R2 given that the polynomials P(x)=x

3

+2x

2

-5ax-7andx

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_15 a-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:1R-1=R-2R

1

=R

2

Answer 2

Answer:

R1=R2

R1+=R2=0

2R1+R2=0