Question

if 2 and -2 are zeroes of the polynomial p(x) = ax4+2x3 -3x2 + bx -4, find the value of a and b

Answer 1

Answer:

a=3/2 and b=-4

Step-by-step explanation:

x=2

ax^4+2x^3 -3x^2 + bx -4=0 => 16a+16-12+2b-4=0 =>16a+2b=16

8a+b=8 ------>(1)

x=-2

ax^4+2x^3 -3x^2 + bx -4=0 => 16a-16-12-2b-4=0 => 16a-2b=32

8a-b=16  ------>(2)

Combining (1) and (2) we find values of a and b:

16a=24=> a=3/2 and b=-4

Answer 2

Given:

2 and -2 are zeroes of the polynomial $p(x) = ax^4+2x^3 -3x^2 + bx -4$

To Find:

Find the value of a and b.

Solution:

For x=2,

    $ax^4+2x^3 -3x^2 + bx -4=0$

⇒ $16a+16-12+2b-4=0$

⇒16a+2b=16

By Simplifying, we get

    8a+b=8                  ------ (1)

For x = -2,

  $ax^4+2x^3 -3x^2 + bx -4=0$

⇒$16a-16-12-2b-4=0$

⇒16a-2b=32

By Simplifying, we get

    8a-b=16        ------(2)

Adding (1) and (2) we find values of a and b:

16a=24=> a=3/2 and b=-4

Hence, values of a and b are 3/2 and -4 respectively.