Factorize #x^3-7x+6#?

2 Answers
Dec 29, 2016

Answer:

Factors of #x^3-7x+6=0# are #(x-1)(x+2)(x-3)=0#

Explanation:

As the coefficients (#1-7+6=0#) in the given equation add up to zero, it is apparent that #x=1# is a solution of #x^3-7x+6=0# and #(x-1)# is a factor of #x^3-7x+6#.

Dividing #x^3-7x+6# by #(x-1)#

#x^3-7x+6=x^2(x-1)+x(x-1)-6(x-1)#

= #(x-1)(x^2+x-6)# - now splitting the middle term

= #(x-1)(x^2+2x-3x-6)#

= #(x-1)(x(x+2)-3(x+2))#

= #(x-1)(x+2)(x-3)#

Hence factors of #x^3-7x+6=0# are #(x-1)(x+2)(x-3)=0#

Dec 29, 2016

Answer:

The answer is #=(x-1)(x+3)(x-2)#

Explanation:

Let #f(x)=x^3-7x+6#

#f(1)=1-7+6=0#

Therefore,

#(x-1)# is a factor

To find the other factors, we do a long division

#color(white)(aaaa)##x^3##color(white)(aaaaa)##-7x+6##color(white)(aaaaaa)##∣##x-1#

#color(white)(aaaa)##x^3-x^2##color(white)(aaaa)####color(white)(aaaaaaaaaa)##∣##x^2+x-6#

#color(white)(aaaaa)##0+x^2-7x#

#color(white)(aaaaaaa)##+x^2-x#

#color(white)(aaaaaaaa)##+0-6x+6#

#color(white)(aaaaaaaaaaaa)##-6x+6#

#color(white)(aaaaaaaaaaaaa)##-0+0#

Therefore,

#(x^3-7x+6)/(x-1)=x^2+x-6=(x+3)(x-2)#

So,

#(x^3-7x+6)=(x-1)(x+3)(x-2)#