Question

find all zeroes of polynomial 2x^4-2x^3-7x^2+3x+6 if its two zeroes are -root 3/2 and root 3/2

Answer 1

two of it zeroes are (x-1) (2x+4)
-1 ,2 ,-√3/2,√3/2 are factors.

Answer 2

Answer:

The remaining zeros are -1 and 2.

Step-by-step explanation:

The given polynomial is

$P(x)=2x^4-2x^3-7x^2+3x+6$

It is given that two zeroes of P(x) are $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$. It means the factors are $(x+\sqrt{\frac{3}{2}})$ and $(x-\sqrt{\frac{3}{2}})$.

$(x+\sqrt{\frac{3}{2}})(x-\sqrt{\frac{3}{2}})=x^2-\frac{3}{2}$

Use long division method to divide P(x) by $x^2-\frac{3}{2}$ to find the remaining factor.

$\frac{2x^4-2x^3-7x^2+3x+6}{x^2-\frac{3}{4}}=2x^2-2x-4$

Equate the remaining factor equal to 0.

$2x^2-2x-4=0$

$2(x^2-x-2)=0$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$x(x-2)+(x-2)=0$

$(x+1)(x-2)=0$

Equate each factor equal to 0.

$x=-1$

$x=2$

Therefore the remaining zeros are -1 and 2.