Factor 2x^2 + 4x.

Step-by-step solution

1. 1

   Build the GCF from the numbers and the variables separately. For the coefficients, $\gcd(2,4)=2$. For the variables, $x^{2}$ and $x$ share one factor of $x$ — always the lowest power present. Multiplying the two parts:

   $\text{GCF}=2x$

   Taking only $2$, or only $x$, would leave the job half done.

2. 2

   Divide each term by $2x$.

   $\frac{2x^{2}}{2x}=x,\qquad\frac{4x}{2x}=2$

   so the quotients $x$ and $2$ form the bracket:

   $2x^{2}+4x=2x(x+2)$

3. 3

   Check the bracket is fully reduced. Inside, $x$ and $2$ share no common factor, so nothing more can come out — the factorisation is complete. (By contrast, the partial answer $2\left(x^{2}+2x\right)$ is a true identity but still hides a common $x$.)

4. 4

   Verify by expanding. $2x\cdot x+2x\cdot2=2x^{2}+4x$ ✓, confirmed at every integer from $-20$ to $19$ ✓.

5. 5

   Read off the roots. Setting $2x(x+2)=0$ and applying the zero-product property gives $x=0$ or $x=-2$. The constant $2$ contributes no root. Both check out: $2(0)+0=0$ ✓ and $2(4)+4(-2)=8-8=0$ ✓.