obtain all the zeros of 3 x power 4 + 6 x cube minus 2 x square - 10 x minus 5 if two of its zeros are root 5 by 3 and minus root by 5 by 3
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Answered by
69
p(x)=3x4+6x3-2x2-10x-5
as we have given that two of its zeroes are root5/3 and -root5/3. so their factors are;
(x-root5/3) and (x+root5/3). the product of the factors ;
(x-root5/3) (x+root5/3)=x2-5/3
=3x2-5/3=1/3(3x2-5)
where k=1/3
now g(x)=3x2-5
dividing p(x) by g(x) we get
q(x)=x2+2x+1
by division algorithm;
p(x)=g(x)*q(x)+r(x)
=(3x2-5)(x2+2x+1)
=(3x2-5)(x2+x+x+1)
=(3x2-5)(x(x+1)+1(x+1)
(3x2-5)(x+1)(x+1)
the other two zeroes are -1 and-1
as we have given that two of its zeroes are root5/3 and -root5/3. so their factors are;
(x-root5/3) and (x+root5/3). the product of the factors ;
(x-root5/3) (x+root5/3)=x2-5/3
=3x2-5/3=1/3(3x2-5)
where k=1/3
now g(x)=3x2-5
dividing p(x) by g(x) we get
q(x)=x2+2x+1
by division algorithm;
p(x)=g(x)*q(x)+r(x)
=(3x2-5)(x2+2x+1)
=(3x2-5)(x2+x+x+1)
=(3x2-5)(x(x+1)+1(x+1)
(3x2-5)(x+1)(x+1)
the other two zeroes are -1 and-1
melvin25:
Is it crt bro
Answered by
20
Answer:
p(x)=3x4+6x3-2x2-10x-5
as we have given that two of its zeroes are root5/3 and -root5/3. so their factors are;
(x-root5/3) and (x+root5/3). the product of the factors ;
(x-root5/3) (x+root5/3)=x2-5/3
=3x2-5/3=1/3(3x2-5)
where k=1/3
now g(x)=3x2-5
dividing p(x) by g(x) we get
q(x)=x2+2x+1
by division algorithm;
p(x)=g(x)*q(x)+r(x)
=(3x2-5)(x2+2x+1)
=(3x2-5)(x2+x+x+1)
=(3x2-5)(x(x+1)+1(x+1)
(3x2-5)(x+1)(x+1)
the other two zeroes are -1 and-1
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