`color{green} ✍️ ` You know that a cubic polynomial has at most three zeroes.
`color{green} ✍️` However, if you are given only one zero, can you find the other two?
`color(red)(=>"For this, let us consider the cubic polynomial")` `color(blue)(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,` as you have
learnt in Class IX, 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
`color(blue)(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`.
`color{green} ✍️ ` You know that a cubic polynomial has at most three zeroes.
`color{green} ✍️` However, if you are given only one zero, can you find the other two?
`color(red)(=>"For this, let us consider the cubic polynomial")` `color(blue)(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,` as you have
learnt in Class IX, 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
`color(blue)(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`.