Question

Find all the zeroes of the polynomial $x^4-8x^3+23x^2-28x+12$ if two of its zeroes are 𝟐 and 𝟑.

Answer 1

Step 1: Find the factor using zeros of the polynomial and find the divisor.

2 and 3 are the zeros of the given polynomial.
Hence, (𝒙 – 2) and (𝒙 – 3) will be the factors of the polynomial.
g(x) = (𝒙 – 2)(𝒙 - 3)
$g\left(x\right)=x^2–5x+6$

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

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

multiply the quotient obtained in the previous step with divisor, hence the product is $x^4-5x^3+6x^2$

Subtract the product of the divisor and the quotient from the dividend.

we get $-2x^2+3x-1$ after subtraction

Repeat these steps till the remainder is zero or deg r(x) < deg g(x).

So, the quotient is $x^2-3x+2$ and the remainder is 0.



Step 3: Factorize the quotient

$x^2-3x+2=0$
$x^2-2x-x+2=0$
$x(x-2)-1(x-2)=0$
$(x-1)(x-2)=0$
$x=1,2$

Hence, the zeroes of the polynomial are 1, 2 and 3.