How do you factor #x^3-49x#?

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Answer 1 · 2018-03-03T17:31:29

Since we have #x^3#, we know that we need one #x# out and two #x# inside the parentheses.

And we know #(7)(7)# is 49. Since it is -49, one of the seven will be negative.

Thus

The answer is: #x(x+7)(x-7)#

Answer 2 · 2018-03-03T18:40:04

#x(x+7)(x-7)#

We can see that both terms contain an #x# which we can factor out to get #x(x^2-49)#

Now we can use difference of two squares to factorise #x^2-49#. Difference of two squares tells us that #a^2-b^2=(a+b)(a-b)#

#x^2-49=(x+7)(x-7)# since #7^2=49#

Substituting this gives us #x(x+7)(x-7)#