Answer :

Answer: True

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x^2 = x*x
5x = 5*x
Both x^2 and 5x have a factor of x in common

x^2 + 5x = x*x + 5*x
x^2 + 5x = x*x + x*5
x^2 + 5x = x*(x+5) .... use the distributive property in reverse

x^2 + 5x factors to x*(x+5)

As a way to check, you can distribute the x back in
x*(x+5) = x*x + x*5
x*(x+5) = x^2 + 5x
and we get the original expression again
budwilkins

[tex] {x}^{2} + 5x \\ x(x + 5)[/tex]
notice there's no whole number in front of the x^2, so no number can factor out both terms, but both have at least one x, so one x factors out of both

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