Question

her zeroes of the polynomial p(x) = 2x4-6x +3x2 +3x-2 if two of zeroes are 1 and 1
2 12​

Answer 1

Answer:

Step-by-step explanation:

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.