Question

find the all zeros of polynomial x^4+x^3-34x^3-4x+120 ,if two of its zeros are 2 and - 2​

Answer 1

Step-by-step explanation:

here is your answer and once try to solve it yourself

Answer 2

Answer:

Step-by-step explanation:Let p(x) = x⁴+x³-34x²-4x+120

Since, two zeroes are 2,-2 ,therefore,

(x-2)(x+2) = x²-2²= x²-4  is a factor of p(x).

Now, we apply the division algorithm to the given p(x) and (x²-4)

(x²-4)x⁴+x³-34x²-4x+120(x²+x-30

*****x⁴ +0- 4x²

__________________

******** x³-30x²-4x

******** x³+ 0 -4x

_____________________

_*********** -30x² + 120

*********** -30x² + 120

______________________

Remainder (0)

So, p(x) = (x²-4)(x²+x-30)Now, x²+x-30

Now,  

x²+x-30

splitting the middle term, we get

= x²+6x-5x-30

= x(x+6)-5(x+6)

= (x+6)(x-5)

Therefore,

p(x) = (x-2)(x+2)(x+6)(x-5)

The other zeroes p(x) are -6 and 5

hope it helps you.... :)