Question

find all other zeros of the polynomial x4 - 2x cube - 7 x square + 8 x + 12 if two of its zeros are - 1 and 2​

Answer 1

Answer:

All the required zeroes of the polynomial are 3 , - 2 , - 1 and 2

Step-by-step explanation:

Given polynomial: $x^{4}$ - 2$x^{3}$ - 7x + 8x + 12

Let one zero of the given polynomial be alpha = - 1

Let the other zero of the given polynomial be beta = 2

Alpha + Beta = 2 - 1 = 1

Alpha * Beta = (2) * (-1) = - 2

Framing quadratic polynomial = $x^{2}$ - (Sum of Zeroes)x + (Product of zeroes)

∴ Quadratic polynomial is $x^{2}$ - (1)x - 2

= $x^{2}$ - x - 2

Dividing:  $x^{4}$ - 2$x^{3}$ - 7x + 8x + 12 / $x^{2}$ - x - 2

After dividing by using long division we get remainder as 0 and quotient as $x^{2}$ - x - 6

q(x) =  $x^{2}$ - x - 6

We can solve this quadratic equation by using factorization:

=> $x^{2}$ - x - 6                            p = - 6    s = - 1 ( Here we have to                               find a number that gives product as - 6 and sum as - 1)

                                                       

=> $x^{2}$ - 3x + 2x - 6  (So, we can take - 3 and 2)

=> x(x - 3) + 2 (x - 3)

=> (x-3) (x + 2)

∴ x = 3 and x = -2

Therefore, all the required zeroes of the polynomial are 3 , - 2 , - 1 and 2