obtain all zeros of the polynomial 2x4-6x3+3x2+3x-2 if two of its zeros are +1 root 2 and -1 root2
Answers
So, polynomial P(x)= 2x^{4}-6x^{3}+3x^{2}+3x-22x
4
−6x
3
+3x
2
+3x−2 and it's zeroes are 1/√2 and -1/√2. We need to find all other zeroes. Let x be the all other zeroes.
(x-1/√2)(x+1/√2)
(x²-1/2)=0
(2x²-1)=0
Hence, the factor is 2x²-1. Therefore,
x^{2} -3x+2x
2
−3x+2
___________________________
2x^{2} -12x
2
−1 2x^{4}-6x^{3}+3x^{2}+3x-22x
4
−6x
3
+3x
2
+3x−2
2x^{4}-x^{2}2x
4
−x
2
___________________________
-6x^{3}+4x^{2}+3x-2−6x
3
+4x
2
+3x−2
-6x^{3}+3x−6x
3
+3x
______________________
4x^{2}-24x
2
−2
4x^{2}-24x
2
−2
_________________
x
Now, 2x^{4}-6x^{3}+3x^{2}+3x-22x
4
−6x
3
+3x
2
+3x−2 = (2x^{2}-1)(x^{2}-3x+2)(2x
2
−1)(x
2
−3x+2)
On future factorizing (x^{2}-3x+2)(x
2
−3x+2) we get,
⇒ x^{2}-2x-x+2=0x
2
−2x−x+2=0
⇒ x(x-2)-1(x-2)=0x(x−2)−1(x−2)=0
⇒ (x-2)(x-1)=0(x−2)(x−1)=0
∴ x=2 and x=1
2 and 1 are other zeroes of the polynomial.