#2x^3-7x^2-2x+7#? How to factor it completely?

factor completely

2 Answers
May 8, 2018

#p(x)=(x+1)(x-1)(2x-7)#

Explanation:

We take,

#p(x)=2x^3-7x^2-2x+7#

#=2x^3-2x-7x^2+7#

#=2xcolor(red)((x^2-1))-7color(red)((x^2-1))#

#=(x^2-1)(2x-7)#

#=(x+1)(x-1)(2x-7)#

May 8, 2018

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

Explanation:

First, use trial and error to find an #x# value that will make the equation equal to zero. From the coefficients, you can tell that #x=1# will make the equation equal to zero.

#2(1)^3 -7(1)^2 -2(1) +7 = 0#

Then you do a polynomial divsion of #(2x^3-7x^2-2x+7)/(x+1)#

(Don't know how to format this so I wrote it out and took a picture of it)

Zero Two

#2x^3-7x^2-2x+7 = (x+1)(2x^2-9x+7)#


(The explaining of polynomial long division)

Now factorise the quadratic.

#2x^2 - 9x+7 = x^2-9x+14#

Which factors add up to nine and their product is 14? The factors are 2 and 7.

# x^2-9x+14=(x-2)(x-7)#

Then put the coefficient of #x# back in and simplify where necessary.

#(2x-2)(2x-7) = (x-1)(x-7)#

#2x^2 -9x +7 = (x-1)(x-7)#

#A= (x+1)(x-1)(2x-7)#