Question

factorise x^3-7x+6 Please please please

Answer 1

Hey mate here's your answer

Answer 2

Answer:  The required factored form of the given expression is $(x-1)(x-2)(x+3).$

Step-by-step explanation:  We are given to factorize the following cubic expression :

$p(x)=x^3-7x+6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Factor theorem :  If x = a is a zero of a polynomial f(x), then (x - a) is a factor of f(x).

Substituting x = 1 in polynomial (i), we have

$p(1)=1^3-7\times1+6=7-7=0.$

So, by factor theorem, (x - 1) is a factor of the polynomial p(x).

From (i), we get

$p(x)\\\\=x^3-7x+6\\\\=x^2(x-1)+x(x-1)-6(x-1)\\\\=(x-1)(x^2+x-6)\\\\=(x-1)(x^2+3x-2x-6)\\\\=(x-1)(x(x+3)-2(x+3))\\\\=(x-1)(x-2)(x+3).$

Thus, the required factored form of the given expression is $(x-1)(x-2)(x+3).$