Question

find the zeros of the polynomial 2x^4 - 10x^3 +5x^2 + 15x - 12 if it is given that the two zeros are root3/2 and -root3/2

Answer 1

p(x)=2x^4-10x³+5x²+15x-12
x= -√3/2 and √3/2
Therefore,
(x+√3/2)(x-√3/2)=0
x²-3/2=0
(2x²-3)/2=0
2x²-3=0
g(x)=2x²-3
Now divide p(x) by g(x)
The q(x) will be
= x²-5x+4
=x²-4x-x+4
=x(x-4)-1(x-4)
=(x-4)(x-1)
x=4 and x=1
Therefore, the other two zeroes are 4 and 1.

Answer 2

Answer:

p(x)=2x^4-10x³+5x²+15x-12

x= -√3/2 and √3/2

Therefore,

(x+√3/2)(x-√3/2)=0

x²-3/2=0

(2x²-3)/2=0

2x²-3=0

g(x)=2x²-3

Now divide p(x) by g(x)

The q(x) will be

= x²-5x+4

=x²-4x-x+4

=x(x-4)-1(x-4)

=(x-4)(x-1)

x=4 and x=1

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