$$a+b=4$$ $$ab=1\times 3=3$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. The only such pair is the system solution.$$a=1$$ $$b=3$$

Rewrite $x^{2}+4x+3$ as $\left(x^{2}+x\right)+\left(3x+3\right)$.$$\left(x^{2}+x\right)+\left(3x+3\right)$$

Factor out $x$ in the first and $3$ in the second group.$$x\left(x+1\right)+3\left(x+1\right)$$

Factor out common term $x+1$ by using distributive property.$$\left(x+1\right)\left(x+3\right)$$

To find equation solutions, solve $x+1=0$ and $x+3=0$.$$x=-1$$ $$x=-3$$

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