You know that a cubic polynomial has at most three zeroes. However, if you are given only one zero, can you find the other two? For this, let us consider the cubic polynomial ${x^3 – 3x^2 – x + 3}$.If we tell you that one of its zeroes is 1, then you know that ${x-1}$ is a factor of ${x^3 – 3x^2 – x + 3}$. So, you can divide ${x^3 – 3x^2 – x + 3}$ by ${x-1}$, to get the quotient ${x^2 – 2x – 3}$.
Next, you could get the factors of ${x^2 – 2x – 3}$, by splitting the middle term, as ${(x + 1)(x – 3)}$. This would give you ${x^3 – 3x^2 – x + 3 = (x – 1)(x^2 – 2x – 3)}$ ${= (x – 1)(x + 1)(x – 3)}$ So, all the three zeroes of the cubic polynomial are now known to you as 1, – 1, 3.

Example: Divide ${2x^2 + 3x + 1}$ by ${x + 2}$.

Steps 1: Take the highest degree value of divisor i.e , in this case ${2x^2}$ and highest degree value of divident i.e in this case ${x}$ and divide them. The quotient shall be ${2x}$

Step 2: Stop the division process when either the remainder is zero or its degree is less than the degree of the divisor.

Step for alorithm division

Solution : Note that we stop the division process when either the remainder is zero or its degree is less than the degree of the divisor. So, here the quotient is ${2x – 1}$ and the remainder is 3. Also,${(2x – 1)(x + 2) + 3}$ = ${2x^2 + 3x – 2 + 3}$ = ${2x^2 + 3x + 1}$ = ${(x + 2)(2x – 1) + 3}$ Therefore, Dividend = Divisor × Quotient + Remainder

Example Find the zeroes of the cubic polynomial $x^3-x^2-4x+4$ if one of the zeroes is 2. Solution Since 2 is one of the zeroes of the polynomial , so $(x-2)$ is a factor of ${x^3-x^2-4x+4}$. So, applying division algorithm,

So, $x^3-x^2-4x+4$ = $(x^2+x-2)(x-2)$ = $(x^2+2x-x-2)(x-2)$ = $(x(x+2)-1(x+2))(x-2)$ = $(x+2)(x-1)(x-2)$ So, zeroes of the cubic polynomial are $1,2 ,-2$