Question

15. The polynomials 2x3 - 7x2 + ax - 6 and
r3 - 8x2 + (2a + 1)x - 16 leave the same
remainder when divided by x - 2. Find the
value of 'a'.

Answer 1

Answer:

Let f(x)=2x

3

−7x

2

+ax−6

Put x−2=0

⇒x=2

When f(x) is divided by (x−2), remainder =f(2)

∴f(2)=2(2)

3

−7(2)

2

+a.2−6

=2.8−7.4+2a−6

=16−28−6+2a

=2a−18

Let g(x)=x

3

−8x

2

+(2a+1)x−16

when g(x) is divided by (x−2) remainder =g(2)

∴g(2)=(2)

3

−8(2)

2

+(2a+1)2−16

=8−32+4a+2−16

=4a−38

By the condition we have

f(2)=g(2)

2a−18=4a−38

4a−2a=38−18

2a=20

a=

2

20

a=10

∴ Thus, the value of a=10

Answer 2

let f(x) = 2x³ - 7x² + ax - 6

put x - 2 = 0

=> x = 2

When f(x) is divided by (x−2), remainder = f(2).

∴ f(2) = 2(2)³ − 7(2)² + a.2 − 6

= 2.8 - 7.4 + 2a - 6

= 16 - 28 - 6 + 2a

= 2a - 18

Let g(x)= x 3 − 8x 2 + (2a+1)x − 16

∴g(2)=(2) 3 −8(2) 2 +(2a+1)2−16

=8−32+4a+2−16

=4a−38

By the condition we have

f(2)=g(2)

2a−18=4a−38

4a−2a=38−18

2a=20

a= 20/2

a=10

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