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 $x(x+7)(x-7)$

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