Question

Find all the zeroes of the polynomial$p\left(x\right)=2x^4-9x^3+5x^2+3x-1$ if two of its zeroes are $2+\surd 3$ and $2-\surd 3$.

Answer 1

Step 1: Find the factor using zeros of the polynomial and find the divisor.
$2+\surd 3$ and $2-\surd 3$ are the zeros of the given polynomial $2x^4-9x^3+5x^2+3x-1$.
Hence, $(x–2-\surd 3)$ and $(x–-2+\surd 3)$ will be the factors of the polynomial
$g\left(x\right)=(x–2-\surd 3)(x-2+\surd 3)$
$g\left(x\right)=(x-2{)}^2-(\surd 3{)}^2$
$g\left(x\right)=x^2–4x+4-3$
$g\left(x\right)=x^2-4x+1$

Step 2: Apply the division Algorithm. Divide the highest degree term of the dividend by the highest degree term of the divisor.

here $2x^4$ is the highest term in dividend and $x^2$ is the highest term in divisor and we will divide $2x^4$ by $x^2$, we get the first term of the quotient as $2x^2$

Step 3: Multiply the quotient with the divisor.

multiply the quotient obtained in the previous step with divisor, hence the product is $2x^4-8x^3+2x^2$

Step 4: Subtract the product of the divisor and the quotient from the dividend.

we get $-x^3+3x^2+3x$
after subtraction

Repeat the steps 2 to 4 till the remainder is zero or deg r(x) < deg g(x).
So, the quotient is $2x^2-x-1$ and the remainder is $0$.

Step 5: Equate the quotient with 0 and factorize the quadratic polynomial to find the remaining zeroes.

$2x^2-x-1=0$
$2x^2-2x+x-1=0$
$2x(x-1)+1(x-1)=0$
$(2x+1)(x-1)=0$
$x=-1/2,1$
So the zeros of the polynomial are $-1/2,1,2+\surd 3$ and $2-\surd 3$.