Question

Find all the zeroes of thr polynomial of x4+x3-34x2-4x+120 if two of iys zeroes are2,-2

Answer 1

hence it's other two zeroes are 5 and -6

Answer 2

Solution:

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

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

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