How do you factor #x ^ { 3} + 5x ^ { 2} - x - 5#?

3 Answers
Jacobi J.
Jul 9, 2018

#(x+5)(x+1)(x-1)#

Explanation:

We have the following:

#color(purple)(x^3+5x^2)color(steelblue)(-x-5)#

We can factor a #x^2# out of the purple terms, and a #-1# out of the blue terms. We will get

#color(purple)(x^2(x+5))color(steelblue)(-1(x+5))#

Since both terms have an #x+5# in common, we can factor that out:

#(x+5)color(red)((x^2-1))#

What I have in red is a classic difference of squares, which can be factored as #(x+1)(x-1)#. We now have:

#(x+5)(x+1)(x-1)#

Hope this helps!

Harish Chandra Rajpoot
Jul 9, 2018

#(x-1)(x+1)(x+5)#

Explanation:

The sum of coefficients of given cubic polynomial: #x^3+5x^2-x-5# is #1+5-1-5=0# hence #x=1# is a root of polynomial i.e. #(x-1)# is a factor of cubic polynomial: #x^3+5x^2-x-5# which can be factorized as follows

#x^3+5x^2-x-5#

#=x^2(x-1)+6x(x-1)+5(x-1)#

#=(x-1)(x^2+6x+5)#

#=(x-1)(x^2+5x+x+5)#

#=(x-1)(x(x+5)+(x+5))#

#=(x-1)((x+5)(x+1))#

#=(x-1)(x+1)(x+5)#

Harish Chandra Rajpoot
Jul 9, 2018

#(x-1)(x+1)(x+5)#

Explanation:

The sum of coefficients of given cubic polynomial: #x^3+5x^2-x-5# is #1+5-1-5=0# hence #x=1# is a root of polynomial i.e. #(x-1)# is a factor of cubic polynomial: #x^3+5x^2-x-5# which can be factorized as follows

#x^3+5x^2-x-5#

#=x^2(x-1)+6x(x-1)+5(x-1)#

#=(x-1)(x^2+6x+5)#

#=(x-1)(x^2+5x+x+5)#

#=(x-1)(x(x+5)+(x+5))#

#=(x-1)((x+5)(x+1))#

#=(x-1)(x+1)(x+5)#

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