Factorize x^3-7x+6?

2 Answers
Shwetank Mauria
Dec 29, 2016

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

Narad T.
Dec 29, 2016

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

color(white)(aaaa)x^3-x^2color(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)

Related questions
See all questions in Factoring Completely
Impact of this question
7069 views around the world
You can reuse this answer
Creative Commons License