How to factor polynomials: x^3 + 125 = (x+5)((x^2)-5x+25)

[colorfulmlee]

alg III

I don't understand how
x^3 + 125 = (x+5)((x^2)-5x+25)
My textbook has this written as the sum of two cubes, but how exactly do you get the first equation to the other equation?

I also don't understand how x^2(x+5)-9x(x+5) = (x^2 - 9)(x+5)
My textbook labels this as the distributive property, but, again, I don't get how it goes from the first to the second

Explanations would be greatly appreciated.

 

[HallsofIvy]

alg III

I don't understand how
x^3 + 125 = (x+5)((x^2)-5x+25)
My textbook has this written as the sum of two cubes, but how exactly do you get the first equation to the other equation?

Your text book apparently expects you to know (Perhaps it was given previously in the textbook?) that, for any a and b,
\(\displaystyle a^3+ b^3= (a+ b)(a^2- ab+ b^2\). Because \(\displaystyle 125= 5^3\), here we have \(\displaystyle x^2+ 5^3\) so using that previous formula, with a= x and b= 5, \(\displaystyle x^3+ 125= x^3+ 5^3= (x+ 5)(x^2- (5)(x)+ 5^2= (x+ 5)(x^2- 5x+ 25)

I also don't understand how x^2(x+5)-9x(x+5) = (x^2 - 9)(x+5)
My textbook labels this as the distributive property, but, again, I don't get how it goes from the first to the second

Do you know what the "distributive property" is? It says that ac+ bc= (a+ b)c. That is, that we can "factor out" a common factor. Here, \(\displaystyle x^2(x+ 5)- 9x(x+ 5)\) has "x+ 5" as a common factor. That is, \(\displaystyle a= x^2\), \(\displaystyle b= -9x\) and \(\displaystyle c= x+ 5\). \(\displaystyle x^2(x+ 5)- 9x(x+ 5)= (x^2- 9x)(x+ 5)\).

Explanations would be greatly appreciated.

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