How do you factor 5x^2- 35x?

Mar 30, 2018

$5 x \left(x - 7\right)$

Explanation:

take out common factors

$\left(1\right) \text{ take " hcf(5,35) =color(red)(5)" out}$

$5 {x}^{2} - 35 x = \textcolor{red}{5} \left({x}^{2} - 7 x\right)$

$\left(2\right) \text{ take "hcf(x^2,x)=color(blue)(x )" out}$

$5 {x}^{2} - 35 x = \textcolor{red}{5} \left({x}^{2} - 7 x\right) = \textcolor{red}{5} \textcolor{b l u e}{x} \left(x - 7\right)$

Mar 30, 2018

$5 x \left(x - 7\right)$

Explanation:

We got:

$5 {x}^{2} - 35 x$

Take the greatest common factor out, which is $5$.

$= 5 \left({x}^{2} - 7 x\right)$

Notice how $x$ still remains in both factors inside, and so we can take it out, and we get:

$= 5 x \left(x - 7\right)$

Mar 30, 2018

$5 x \left(x - 7\right)$

Explanation:

Both terms are divisible by $5$, and $x$, so you factor those out of the polynomial. As for how this affects any solutions of this, because both factors equal $0$ (or whatever is on the other side of the equation, but I'll use $0$), $5 x = 0$, so $x = 0$, and $x - 7 = 0$, so $x = 7$