Factoring Cubic Polynomials

[Zahrticus]

I desperately need some clarity on this. The equation is factored out to x^2(2x+5)-(2x+5). That makes some sense to me but where I really get lost is when it says to factor out (2x+5) from each term. This is where they lose me because I don't understand how you are left with (2x+5)(x^2-1). My brain says get rid of both of the instances of (2x+5) and you are left only with (x^2-1). Can you help me understand this please? Thanks in advance.

 

[Romsek]

Not get rid of both incidences of (2x+5), but to group them together.

Their wording is poor. They mean factor (2x+5) from each term, and then group those terms together as a single term multiplying (2x+5)

 

[Zahrticus]

Clarity in math is so satisfying! Thank you so much, that makes perfect sense and was extremely helpful!

 

[Steven G]

The equation is factored out to x^2(2x+5)-(2x+5) This is NOT factored as there are two terms (they are each underlined). An expression that is factored only has one term.
1st notice that x^2(2x+5)-(2x+5) = x^2(2x+5)-1(2x+5) = x^2(2x+5)-1(2x+5). Note that (2x+5) is in each term. So when we factor, one term will be (2x+5) and the other term will be literally what is left over. That is the 2nd term will be exactly x^2(2x+5)-1(2x+5) except there will be no (2x+5), so it will be (x^2-1).
Hence x^2(2x+5)-(2x+5) = (2x+5)(x^2-1). Now is this completely factored??

 

[HallsofIvy]

The "distributive law": ac+ bc= (a+ b)c. Here \(\displaystyle a= x^2\), b= -1. and \(\displaystyle c= 2x+ 5\).