Question

13. Define factor theorem and factorise x3-1​

Answer 1

Answer:

It is a theorem that links factors and zeros of the polynomial. According to factor theorem, if f(x) is a polynomial of degree n ≥ 1 and 'a' is any real number, then, (x-a) is a factor of f(x), if f(a)=0. Also, we can say, if (x-a) is a factor of polynomial f(x), then f(a) = 0. This proves the converse of the theorem.

x3 -1 = ( x - 1 ) (x^2 + x + 1)

Answer 2

Given:

Define factor theorem and factorise $\[{x^3} - 1\]$,

Solution:

Know that, the factor theorem states that if $f(x)$ is a polynomial of degree$n ≥ 1$and 'a' is any real number, then, $(x-a)$ is a factor of $\[f\left( x \right),{\rm{ if }}f\left( a \right) = 0.\]$

Factorise  $\[{x^3} - 1\]$,

Use algebraic identity, $\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]$

Therefore,

$\[\begin{array}{l}{x^3} - 1 = {x^3} - {1^3}\\\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)\end{array}\]$

Hence, the factorisation of  $\[{x^3} - 1\]$ is $\[\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)\]$.