Question

1. If 2x^3 ax^2+bx-6 has (x-1) as a factor and leaves a remainder 2 when divided by (x-2) find them values of a and b

Answer 1

Answer:

$a=-8,b=12$

Step-by-step explanation:

The given polynomial is

$p(x)=2x^3+ax^2+bx-6$

Remainder theorem: If a polynomial P(x) is divided by (x-c), then the number remainder is P(c).

It is given that (x-1) is a factor of given polynomial. It means the remainder is 0 when polynomial is divided by (x-1).

By using Remainder theorem, we get

$p(1)=0$

$2(1)^3+a(1)^2+b(1)-6=0$

$2+a+b-6=0$

$a+b-4=0$

$a+b=4$                  .... (1)

It leaves a remainder 2 when it is divided by (x-2).

By using Remainder theorem, we get

$p(2)=2$

$2(2)^3+a(2)^2+b(2)-6=2$

$16+4a+2b-6=2$

$4a+2b+10=2$

$4a+2b=2-10$

$4a+2b=-8$              .... (2)

Substitute the value of a in equation (2) from equation (1).

$4(4-b)+2b=-8$

$16-4b+2b=-8$

$16-2b=-8$

Subtract 16 from both sides.

$-2b=-8-16$

$-2b=-24$

Divide both sides by -2.

$b=12$

The value of b is 12. Substitute the value of b is equation (1).

$a+12=4$  

Subtract 12 from both sides.

$a=4-12$  

$a=-8$

Therefore, the values of a and b are -8 and 12 respectively.

Answer 2

Step-by-step explanation:

The given polynomial is

p(x)=2x^3+ax^2+bx-6p(x)=2x3+ax2+bx−6

Remainder theorem: If a polynomial P(x) is divided by (x-c), then the number remainder is P(c).

It is given that (x-1) is a factor of given polynomial. It means the remainder is 0 when polynomial is divided by (x-1).

By using Remainder theorem, we get

p(1)=0p(1)=0

2(1)^3+a(1)^2+b(1)-6=02(1)3+a(1)2+b(1)−6=0

2+a+b-6=02+a+b−6=0

a+b-4=0a+b−4=0

a+b=4a+b=4                  .... (1)

It leaves a remainder 2 when it is divided by (x-2).

By using Remainder theorem, we get

p(2)=2p(2)=2

2(2)^3+a(2)^2+b(2)-6=22(2)3+a(2)2+b(2)−6=2

16+4a+2b-6=216+4a+2b−6=2

4a+2b+10=24a+2b+10=2

4a+2b=2-104a+2b=2−10

4a+2b=-84a+2b=−8              .... (2)

Substitute the value of a in equation (2) from equation (1).

4(4-b)+2b=-84(4−b)+2b=−8

16-4b+2b=-816−4b+2b=−8

16-2b=-816−2b=−8

Subtract 16 from both sides.

-2b=-8-16−2b=−8−16

-2b=-24−2b=−24

Divide both sides by -2.

b=12b=12

The value of b is 12. Substitute the value of b is equation (1).

a+12=4a+12=4  

Subtract 12 from both sides.

a=4-12a=4−12  

a=-8a=−8

Therefore, the values of a and b are -8 and 12 respectively.