Question

if 2x³+ax²+bx-2 has a factor (x+2) and leaves a reminder 7 when divided by (2x-3) find the value of a and b ​

Answer 1

Let P(x) = 2x3 + ax2 + bx – 2

when P(x) is divided by 2x – 3

P(3/2) = 2(3/2)3 + a(3/2)2 + b(3/2) – 2 = 7

= 27/4 + 9/4a + 3/2b – 2 = 7

= 27 + 9a + 6b – 8/4 = 7

= 9a + 6b = 28 + 8 – 27

= 9a + 6b = 9

⇒ 3a + 2b = 3 ….(1)

Similarly when P(x) is divided by x + 2

x = - 2

2(- 2)3 + a(- 2)2 + b(- 2) – 2 = 0

-16 + 4a – 2b – 2 = 0

⇒ 4a – 2b = 18 ….(2)

On solving equation (1) and (2)



On substituting value of a in equation (1)

3 3 + 2b = 3

2b = 3 – 9

b = -6/2 = - 3

b = - 3

a = 3, b = - 3

On substituting value of a and b

2x3 + 3a2 – 3x – 2

When x + 2 is factor

Answer 2

Answer :

• a = 3
• b = -3

Step-by-step explanation :

Given polynomial => 2x³ + ax² + bx - 2

• (x + 2) is a factor

   => x + 2 = 0

         x = -2

When we substitute x = -2, the result is 0.

Put x = -2,

  2(-2)³ + a(-2)² + b(-2) - 2 = 0

  2(-8) + a(4) - 2b - 2 = 0

    -16 + 4a - 2b - 2 = 0

          4a - 2b = 16 + 2

          4a - 2b = 18

         2(2a - b) = 2(9)

           2a - b = 9  ----[1]

• when divided by (2x - 3). the remainder is 7.

          2x - 3 =0

            2x = 3

             x = 3/2

Substitute x = 3/2 , the result is 7

   $2(\frac{3}{2})^3+a(\frac{3}{2})^2+b(\frac{3}{2})-2=7 \\\\ 2(\frac{27}{8})+a(\frac{9}{4})+\frac{3b}{2}-2=7 \\\\ \frac{27}{4} +\frac{9a}{4}+\frac{3b}{2}=7+2 \\\\ \frac{27+9a+6b}{4} =9 \\\\ 27+9a+6b=9 \times 4 \\\\ 27+9a+6b=36 \\\\ 9a+6b=36-27 \\\\ 9a+6b=9 \\\\ 3(3a+2b) =3(3) \\\\ \bf 3a+2b=3 ---[2]$

from equation [1]

2a - b = 9

 b = 2a - 9

substitute it in equation [2]

3a + 2b = 3

3a + 2(2a - 9) = 3

3a + 4a - 18 = 3

 7a = 18 + 3

  7a = 21

  a = 21/7

   a = 3

=> b = 2a - 9

    b = 2(3) - 9

    b = 6 - 9

    b = -3

∴ a = 3 , b = -3